Chapter 9.1 - Solved Examples On Sequences And Series

In the previous section, we saw the basics about sequences and series. In this section, we will see some solved examples.

Solved example 9.1
(i) Write the first three terms of the sequence defined by the formula: $a_n=2n+5$
(ii) Write the first three terms of the sequence defined by the formula: $a_n=\frac{n-3}{4}$
Solution:
We can find the first three terms by substituting n = 1, 2 and 3  
Part (i):
The first three terms are:
$a_1~=~(2 × 1)+5~=~7$
$a_2~=~(2 × 2)+5~=~9$
$a_3~=~(2 × 3)+5~=~11$
Part (ii):
The first three terms are:
$a_1~=~\frac{1-3}{4}~=~\frac{-2}{4}~=~-\frac{1}{2}$
$a_2~=~\frac{2-3}{4}~=~\frac{-1}{4}~=~-\frac{1}{4}$
$a_3~=~\frac{3-3}{4}~=~\frac{0}{4}~=~0$

Solved example 9.2
What is the 20^th term of the sequence defined by $a_n=(n-1)(2-n)(3+n)$
Solution:
We can find the 20^th term by substituting n = 20
So we get: $a_{20}~=~(20-1)(2-20)(3+20)~=~19 × -18 × 23~=~-7866$

Solved example 9.3
Let the sequence a_n be defined as follows:
$a_1=1,~a_n=a_{n-1}+2~~\text{for}~n \geq 2$
Find the first five terms and write the corresponding series.
Solution:
1. The first five terms are:
$a_1~=~1$
$a_2=a_{2-1}+2~=~a_1+2~=~1+2~=~3$
$a_3=a_{3-1}+2~=~a_2+2~=~3+2~=~5$
$a_4=a_{4-1}+2~=~a_3+2~=~5+2~=~7$
$a_5=a_{5-1}+2~=~a_4+2~=~7+2~=~9$
2. So the sequence is: 1, 3, 5, 7, 9, . . .
3. So the series associated with this sequence is:
1 + 3 + 5 + 7 + 9 + . . .

---

Exercise 9.1

1. Write the first five terms of the sequence whose n^th term is $a_n=n(n+2)$
Solution:
$\begin{array}{ll}
{a_1}&{}={}
&{1 × (1+2)}& {}={}
&3& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{a_2}&{}={}
&{2 × (2+2)}& {}={}
&8& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{a_3}&{}={}
&{3 × (3+2)}& {}={}
&15& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{a_4}&{}={}
&{4 × (4+2)}& {}={}
&24& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{a_5}&{}={}
&{5 × (5+2)}& {}={}
&35& {}
&{}& {}
&{}& {}&{}& {} &{} \\

\end{array}$

2. Write the first five terms of the sequence whose n^th term is $a_n=\frac{n}{n+1}$
Solution:
$\begin{array}{ll}
{a_1}&{}={}
&{\frac{1}{1+1}}& {}={}
&\frac{1}{2}& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{a_2}&{}={}
&{\frac{2}{2+1}}& {}={}
&\frac{2}{3}& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{a_3}&{}={}
&{\frac{3}{3+1}}& {}={}
&\frac{3}{4}& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{a_4}&{}={}
&{\frac{4}{4+1}}& {}={}
&\frac{4}{5}& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{a_5}&{}={}
&{\frac{5}{5+1}}& {}={}
&\frac{5}{6}& {}
&{}& {}
&{}& {}&{}& {} &{} \\

\end{array}$

3. Write the first five terms of the sequence whose n^th term is $a_n=2^n$
Solution:
$\begin{array}{ll}
{a_1}&{}={}
&{2^1}& {}={}
&2& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{a_2}&{}={}
&{2^2}& {}={}
&4& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{a_3}&{}={}
&{2^3}& {}={}
&8& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{a_4}&{}={}
&{2^4}& {}={}
&16& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{a_5}&{}={}
&{2^5}& {}={}
&32& {}
&{}& {}
&{}& {}&{}& {} &{} \\

\end{array}$

4. Write the first five terms of the sequence whose n^th term is $a_n=\frac{2n-3}{6}$
Solution:
$\begin{array}{ll}
{a_1}&{}={}
&{\frac{(2 × 1)-3}{6}}& {}={}
&\frac{2-3}{6}& {}={}
&{-\frac{1}{6}}& {}
&{}& {}&{}& {} &{} \\

{a_2}&{}={}
&{\frac{(2 × 2)-3}{6}}& {}={}
&\frac{4-3}{6}& {}={}
&{\frac{1}{6}}& {}
&{}& {}&{}& {} &{} \\

{a_3}&{}={}
&{\frac{(2 × 3)-3}{6}}& {}={}
&\frac{6-3}{6}& {}={}
&{\frac{3}{6}}& {}
&{}& {}&{}& {} &{} \\

{a_4}&{}={}
&{\frac{(2 × 4)-3}{6}}& {}={}
&\frac{8-3}{6}& {}={}
&{\frac{5}{6}}& {}
&{}& {}&{}& {} &{} \\

{a_5}&{}={}
&{\frac{(2 × 5)-3}{6}}& {}={}
&\frac{10-3}{6}& {}={}
&{\frac{7}{6}}& {}
&{}& {}&{}& {} &{} \\

\end{array}$

5. Write the first five terms of the sequence whose n^th term is $a_n=(-1)^{n-1}~5^{n+1}$
Solution:
$\begin{array}{ll}
{a_1}&{}={}
&{(-1)^{1-1} × 5^{1+1}}& {}={}
&{(-1)^{0} × 5^{2}}& {}={}
&{1 × 25}& {}={}
&{25}& {}&{}& {} &{} \\

{a_2}&{}={}
&{(-1)^{2-1} × 5^{2+1}}& {}={}
&{(-1)^{1} × 5^{3}}& {}={}
&{-1 × 125}& {}={}
&{-125}& {}&{}& {} &{} \\

{a_3}&{}={}
&{(-1)^{3-1} × 5^{3+1}}& {}={}
&{(-1)^{2} × 5^{4}}& {}={}
&{1 × 625}& {}={}
&{625}& {}&{}& {} &{} \\

{a_4}&{}={}
&{(-1)^{4-1} × 5^{4+1}}& {}={}
&{(-1)^{3} × 5^{5}}& {}={}
&{-1 × 3125}& {}={}
&{-3125}& {}&{}& {} &{} \\

{a_5}&{}={}
&{(-1)^{5-1} × 5^{5+1}}& {}={}
&{(-1)^{4} × 5^{6}}& {}={}
&{1 × 15625}& {}={}
&{15625}& {}&{}& {} &{} \\

\end{array}$

6. Write the first five terms of the sequence whose n^th term is $a_n=n\frac{n^2+5}{4}$
Solution:
$\begin{array}{ll}
{a_1}&{}={}
&{1 × \frac{1^2+5}{4}}& {}={}
&{1 × \frac{1+5}{4}}& {}={}
&{1 × \frac{6}{4}}& {}={}
&{\frac{6}{4}}& {}&{}& {} &{} \\

{a_2}&{}={}
&{2 × \frac{2^2+5}{4}}& {}={}
&{2 × \frac{4+5}{4}}& {}={}
&{2 × \frac{9}{4}}& {}={}
&{\frac{9}{2}}& {}&{}& {} &{} \\

{a_3}&{}={}
&{3 × \frac{3^2+5}{4}}& {}={}
&{3 × \frac{9+5}{4}}& {}={}
&{3 × \frac{14}{4}}& {}={}
&{\frac{42}{4}}& {}&{}& {} &{} \\

{a_4}&{}={}
&{4 × \frac{4^2+5}{4}}& {}={}
&{4 × \frac{16+5}{4}}& {}={}
&{4 × \frac{21}{4}}& {}={}
&{21}& {}&{}& {} &{} \\

{a_5}&{}={}
&{5 × \frac{5^2+5}{4}}& {}={}
&{5 × \frac{25+5}{4}}& {}={}
&{5 × \frac{30}{4}}& {}={}
&{\frac{150}{4}}& {}&{}& {} &{} \\

\end{array}$

7. Find a₁₇ and a₂₄ of the sequence whose n^th term is $a_n=4n-3$
Solution:
$\begin{array}{ll}
{a_{17}}&{}={}
&{(4 × 17)-3}& {}={}
&{68-3}& {}={}
&{65}& {}
&{}& {}&{}& {} &{} \\

{a_{24}}&{}={}
&{(4 × 24)-3}& {}={}
&{96-3}& {}={}
&{93}& {}
&{}& {}&{}& {} &{} \\

\end{array}$

8. Find a₇ of the sequence whose n^th term is $a_n=\frac{n^2}{2^n}$
Solution:
$\begin{array}{ll}
{a_{7}}&{}={}
&{\frac{7^2}{2^7}}& {}={}
&{\frac{49}{128}}& {}
&{}& {}
&{}& {}&{}& {} &{} \\

\end{array}$  

9. Find a₉ of the sequence whose n^th term is $a_n=(-1)^{n-1} n^3$
Solution:
$\begin{array}{ll}
{a_{9}}&{}={}
&{(-1)^{9-1} × 9^3}& {}={}
&{(-1)^{8} × 729}& {}={}
&{1 × 729}& {}={}
&{729}& {}&{}& {} &{} \\

\end{array}$

10. Find a₂₀ of the sequence whose n^th term is $a_n=\frac{n(n-2)}{n+3}$
Solution:
$\begin{array}{ll}
{a_{20}}&{}={}
&{\frac{20(20-2)}{20+3}}& {}={}
&{\frac{20 × 18}{23}}& {}={}
&{\frac{360}{23}}& {}
&{}& {}&{}& {} &{} \\

\end{array}$

11. Write the first five terms of the following sequence and obtain the corresponding series.
$a_1=3,~a_n=3a_{n-1}+2~~\text{for all}~n \gt 1$
Solution:
1. The first five terms are:
$\begin{array}{ll}
{a_{1}}&{}={}
&{3}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{a_{2}}&{}={}
&{3 × a_{2-1}+2}& {}={}
&{3 × a_{1}+2}& {}={}
&{(3 × 3)+2}& {}={}
&{9+2}& {}={}&{11}& {} &{} \\

{a_{3}}&{}={}
&{3 × a_{3-1}+2}& {}={}
&{3 × a_{2}+2}& {}={}
&{(3 × 11)+2}& {}={}
&{33+2}& {}={}&{35}& {} &{} \\

{a_{4}}&{}={}
&{3 × a_{4-1}+2}& {}={}
&{3 × a_{4}+2}& {}={}
&{(3 × 35)+2}& {}={}
&{105+2}& {}={}&{107}& {} &{} \\

{a_{5}}&{}={}
&{3 × a_{5-1}+2}& {}={}
&{3 × a_{4}+2}& {}={}
&{(3 × 107)+2}& {}={}
&{321+2}& {}={}&{323}& {} &{} \\

\end{array}$

2. So the sequence is: 3, 11, 35, 107, 323, . . .
3. So the series associated with this sequence is:
3 + 11 + 35 + 107 + 323, . . .

12. Write the first five terms of the following sequence and obtain the corresponding series.
$a_1=-1,~a_n=\frac{a_{n-1}}{n}~~\text{for all}~n \ge 2$
Solution:
1. The first five terms are:
$\begin{array}{ll}
{a_{1}}&{}={}
&{-1}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{a_{2}}&{}={}
&{\frac{a_{2-1}}{2}}& {}={}
&{\frac{a_{1}}{2}}& {}={}
&{\frac{-1}{2}}& {}
&{}& {}&{}& {} &{} \\

{a_{3}}&{}={}
&{\frac{a_{3-1}}{3}}& {}={}
&{\frac{a_{2}}{3}}& {}={}
&{\frac{\frac{-1}{2}}{3}}& {}={}
&{\frac{-1}{6}}& {}&{}& {} &{} \\

{a_{4}}&{}={}
&{\frac{a_{4-1}}{4}}& {}={}
&{\frac{a_{3}}{4}}& {}={}
&{\frac{\frac{-1}{6}}{4}}& {}={}
&{\frac{-1}{24}}& {}&{}& {} &{} \\

{a_{5}}&{}={}
&{\frac{a_{5-1}}{5}}& {}={}
&{\frac{a_{4}}{5}}& {}={}
&{\frac{\frac{-1}{24}}{5}}& {}={}
&{\frac{-1}{120}}& {}&{}& {} &{} \\

\end{array}$

2. So the sequence is: $-1, \frac{-1}{2}, \frac{-1}{6}, \frac{-1}{24}, \frac{-1}{120}, ~.~.~.$
3. So the series associated with this sequence is:
$-1 + \frac{-1}{2} + \frac{-1}{6} + \frac{-1}{24} + \frac{-1}{120} ~.~.~.$

13. Write the first five terms of the following sequence and obtain the corresponding series.
$a_1=a_2=2,~a_n=a_{n-1}-1~~\text{for all}~n \gt 2$
Solution:
1. The first five terms are:
$\begin{array}{ll}
{a_{1}}&{}={}
&{2}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{a_{2}}&{}={}
&{2}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{a_{3}}&{}={}
&{a_{3-1}-1}& {}={}
&{a_{2}-1}& {}={}
&{2-1}& {}={}
&{1}& {}&{}& {} &{} \\

{a_{4}}&{}={}
&{a_{4-1}-1}& {}={}
&{a_{3}-1}& {}={}
&{1-1}& {}={}
&{0}& {}&{}& {} &{} \\

{a_{5}}&{}={}
&{a_{5-1}-1}& {}={}
&{a_{4}-1}& {}={}
&{0-1}& {}={}
&{-1}& {}&{}& {} &{} \\

\end{array}$

2. So the sequence is: 2, 2, 1, 0, -1, . . .
3. So the series associated with this sequence is:
2 + 2 + 1 + 0 + (-1), . . .

14. The Fibonacci sequence is defined by:
$a_1=a_2=1,~a_n=a_{n-1}~+~a_{n-2}~~\text{for all}~n \gt 2$
Find $\frac{a_{n+1}}{a_n}$ for n = 1, 2, 3, 4, 5
Solution:
1. The first six terms are:
$\begin{array}{ll}
{a_{1}}&{}={}
&{1}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{a_{2}}&{}={}
&{1}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{a_{3}}&{}={}
&{a_{3-1}~+~a_{3-2}}& {}={}
&{a_{2}~+~a_{1}}& {}={}
&{1+1}& {}={}
&{2}& {}&{}& {} &{} \\

{a_{4}}&{}={}
&{a_{4-1}~+~a_{4-2}}& {}={}
&{a_{3}~+~a_{2}}& {}={}
&{2+1}& {}={}
&{3}& {}&{}& {} &{} \\

{a_{5}}&{}={}
&{a_{5-1}~+~a_{5-2}}& {}={}
&{a_{4}~+~a_{3}}& {}={}
&{3+2}& {}={}
&{5}& {}&{}& {} &{} \\

{a_{6}}&{}={}
&{a_{6-1}~+~a_{6-2}}& {}={}
&{a_{5}~+~a_{4}}& {}={}
&{5+3}& {}={}
&{8}& {}&{}& {} &{} \\

\end{array}$

2. So the required ratios are:
$\begin{array}{ll}
{\text{When n = 1,}~\frac{a_{n+1}}{a_n}}&{}={}
&{\frac{a_{1+1}}{a_1}}& {}={}
&{\frac{a_{2}}{a_1}}& {}={}
&{\frac{1}{1}}& {}
&{}& {}&{}& {} &{} \\

{\text{When n = 2,}~\frac{a_{n+1}}{a_n}}&{}={}
&{\frac{a_{2+1}}{a_2}}& {}={}
&{\frac{a_{3}}{a_2}}& {}={}
&{\frac{2}{1}}& {}
&{}& {}&{}& {} &{} \\

{\text{When n = 3,}~\frac{a_{n+1}}{a_n}}&{}={}
&{\frac{a_{3+1}}{a_3}}& {}={}
&{\frac{a_{4}}{a_3}}& {}={}
&{\frac{3}{2}}& {}
&{}& {}&{}& {} &{} \\

{\text{When n = 4,}~\frac{a_{n+1}}{a_n}}&{}={}
&{\frac{a_{4+1}}{a_4}}& {}={}
&{\frac{a_{5}}{a_4}}& {}={}
&{\frac{5}{3}}& {}
&{}& {}&{}& {} &{} \\

{\text{When n = 5,}~\frac{a_{n+1}}{a_n}}&{}={}
&{\frac{a_{5+1}}{a_5}}& {}={}
&{\frac{a_{6}}{a_5}}& {}={}
&{\frac{8}{5}}& {}
&{}& {}&{}& {} &{} \\

\end{array}$

---

In the next section we will see arithmetic progression.

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Chapter 9 - Sequences And Series

In the previous section, we completed a discussion on binomial theorem. In this  chapter, we will see sequences and series.

The word sequence can be explained using three examples:
Example 1:
This can be written in 4 steps:
1. Consider a bundle of freshly printed currency notes.
2. Let there be a hundred notes in the bundle.
• Each note in that bundle will have a unique number.
3. Suppose that, the 14^th note from the top has the number 175214
    ♦ Then the 15^th note from the top will have the number 175215
    ♦ The 16^th note from the top will have the number 175216
    ♦ so on . . .
4. We say that, the currency notes in the bundle are arranged in a sequential order.
• The sequence can be written as:
175201, 175202, 175203, . . . , 175214, 175215, 175216, 175217, . . . , 175299, 175300.

Example 2
This can be written in 3 steps:
1. Suppose that, we want to make equilateral triangles with identical spheres. Then the arrangement will be as shown in fig.9.1 below:

Fig.9.1

2. The first triangle will contain 3 spheres
    ♦ The second triangle will contain 6 spheres
    ♦ The third triangle will contain 10 spheres
    ♦ The fourth triangle will contain 15 spheres
    ♦ so on . . .
3. This is the only way to form equilateral triangles using identical spheres.
• So the number of spheres in the triangles form a sequence.
• The sequence can be written as: 3, 6, 10, 15, . . .

Example 3
• Suppose that in a society, the generation gap is 30 years. That is., a person will have children when he/she is 30 years old.
• We want to find the total number of members of a family in each generation for 300 years.
• This can be calculated in 13 steps:
1. Let the total number of members in 300 years be x.
2. Consider a person whose age at present is 30.
• Then 30 years ago, he was not born. His family had upto his father and mother only. So 30 years ago, the number of members will be (x-1)
3. 60 years ago, his father and mother were not born.
• So 60 years ago, the number of members would be [(x-1)-2] = (x-3)
4. 90 years ago, the parents of the persons deducted in (3) were not born.
• Two persons were deducted in (3).
• There would be four parents for those two persons.
• So 90 years ago, the number of members would be [(x-3)-4] = (x-7)
5. 120 years ago, the parents of the persons deducted in (4) were not born.
• 4 persons were deducted in (4).
• There would be 8 parents for those 4 persons.
• So 120 years ago, the number of members would be [(x-7)-8] = (x-15)  
6. 150 years ago, the parents of the persons deducted in (5) were not born.
• 8 persons were deducted in (5).
• There would be 16 parents for those 8 persons.
• So 150 years ago, the number of members would be [(x-15)-16] = (x-31) 
7. 180 years ago, the parents of the persons deducted in (6) were not born.
• 16 persons were deducted in (5).
• There would be 32 parents for those 16 persons.
• So 180 years ago, the number of members would be [(x-31)-32] = (x-63) 
8. 210 years ago, the parents of the persons deducted in (7) were not born.
• 32 persons were deducted in (7).
• There would be 64 parents for those 32 persons.
• So 210 years ago, the number of members would be [(x-63)-64] = (x-127)
9. 240 years ago, the parents of the persons deducted in (8) were not born.
• 64 persons were deducted in (8).
• There would be 128 parents for those 64 persons.
• So 240 years ago, the number of members would be [(x-127)-128] = (x-255)
10. 270 years ago, the parents of the persons deducted in (9) were not born.
• 128 persons were deducted in (9).
• There would be 256 parents for those 128 persons.
• So 270 years ago, the number of members would be [(x-255)-256] = (x-511)
11. 300 years ago, the parents of the persons deducted in (10) were not born.
• 256 persons were deducted in (10).
• There would be 512 parents for those 256 persons.
• So 300 years ago, the number of members would be [(x-511)-512] = (x-1023)
12. But 300 years ago, there will be just two members. A father and mother.
• So we can write: x-1023 = 2
• So x = 1025
• That is., total numbers of members in 300 years = 1025
• We can write: The present person under consideration has 1024 ancestors.
13. Now consider the number of parents, grand parents, great grand parents etc.,
• We get:
    ♦ Number of parents = 2
    ♦ Number of grand parents = 4
    ♦ Number of great grand parents = 8
    ♦ so on . . .
• These numbers form a sequence. The sequence is:
2, 4, 8, 16, 32, . . .

---

• Let us see some features related to sequences. They can be written in steps:
1. The various numbers occurring in a sequence are called terms of the sequence.
• For example, we can write:
‘8’ is a term of the sequence 2, 4, 8, 16, 32, . . .
2. The terms are denoted by the letter a.
• A subscript is also given to ‘a’. The subscript will indicate the position of the term.
• For example, we can write:
In the sequence 2, 4, 8, 16, 32, . . . , a₄ = 16
3. a_n denotes the term at the n^th position in the sequence. This term is also called the general term of the sequence.
4. If the number of terms in a sequence is finite, then that sequence is called a finite sequence.
• For example, the sequence in example 1 that we saw above, is a finite sequence. This is because, the number of terms in that sequence is 100, which is a finite number. (Something that is finite has a definite fixed size or extent)
• The sequence in example 3 is also finite because, we are considering the number of generations within 300 years.
5. If the number of terms in a sequence is infinite, then that sequence is called a infinite sequence.
• For example, the sequence in example 2 that we saw above, is an infinite sequence. This is because, infinite number of such triangles are possible and correspondingly, infinite number of terms will be present in that sequence. (Something that is infinite, is limitless or endless in space, extent, or size. It is impossible to count, measure or calculate an infinite quantity)

---

Algebraic formula for the n^th term of a sequence

This can be explained with the help of some examples.
Example 1:
This can be written in 4 steps:
1. Consider the sequence 2, 4, 6, 8, . . .
• It is the sequence of even natural numbers.
2. We have:
• a₁ = 2
    ♦ 2 can be obtained by multiplying the ‘position number 1’ by 2
• a₂ = 4
    ♦ 4 can be obtained by multiplying the ‘position number 2’ by 2
• a₃ = 6
    ♦ 6 can be obtained by multiplying the ‘position number 3’ by 2
• a₄ = 8
    ♦ 8 can be obtained by multiplying the ‘position number 4’ by 2
so on . . .
3. So it is clear that, any term in this sequence can be obtained by multiplying it’s ‘position number’ by 2
That is., a_n = n × 2
4. We can write:
The algebraic formula for the n^th term of the sequence 2, 4, 6, 8, . . . is: a_n = 2n

Example 2:
This can be written in 4 steps:
1. Consider the sequence 1, 3, 5, 7, . . .
• It is the sequence of odd natural numbers.
2. We have:
• a₁ = 1
    ♦ 1 can be obtained by multiplying the ‘position number 1’ by 2 and then subtracting 1
    ♦ That is., 1 = (1 × 2) - 1
• a₂ = 3
    ♦ 3 can be obtained by multiplying the ‘position number 2’ by 2 and then subtracting 1
    ♦ That is., 3 = (2 × 2) - 1
• a₃ = 5
    ♦ 1 can be obtained by multiplying the ‘position number 3’ by 2 and then subtracting 1
    ♦ That is., 5 = (3 × 2) - 1
• a₄ = 7
    ♦ 7 can be obtained by multiplying the ‘position number 4’ by 2 and then subtracting 1
    ♦ That is., 7 = (4 × 2) - 1
so on . . .
3. So it is clear that, any term in this sequence can be obtained by multiplying it’s ‘position number’ by 2 and then subtracting 1
That is., a_n = n × 2 - 1
4. We can write:
The algebraic formula for the n^th term of the sequence 1, 3, 5, 7, . . . is:
a_n = 2n - 1  

Example 3:
This can be written in 5 steps:
1. Consider the sequence 1, 1, 2, 3, 5, 8, . . .
• Here we do not see a visible pattern.
2. We have:
• a₁ = 1
• a₂ = 1
• a₃ = 2
    ♦ 2 can be obtained by adding a₂ and a₁.
    ♦ a₂ and a₁ are the two terms coming just before a₃
• a₄ = 3
    ♦ 3 can be obtained by adding a₃ and a₂.
    ♦ a₃ and a₂ are the two terms coming just before a₄.
• a₅ = 5
    ♦ 5 can be obtained by adding a₄ and a₃.
    ♦ a₄ and a₃ are the two terms coming just before a₅.
• a₆ = 8
    ♦ 8 can be obtained by adding a₅ and a₄.
    ♦ a₅ and a₄ are the two terms coming just before a₆.
so on . . .
3. So it is clear that, any term in this sequence can be obtained by adding the two terms coming just before it.
That is., a_n = a_n-1 + a_n-2.
4. We can write:
The algebraic formula for the n^th term of the sequence 1, 1, 2, 3, 5, 8, . . . is:
a_n = a_n-1 + a_n-2, n > 2
• Note that, n must be greater than 2. If we put n = 2, then a_n-2 will denote the zero^th term, which is not available.
5. This sequence is called Fibonacci sequence.

---

• Now we have a basic idea about the algebraic formula for the n^th term.
• In the formula, we put natural numbers starting from 1 on the left side. We get corresponding values on the right side.
• Based on this information, we can consider a sequence as a function of the form a(n)
• In some cases, the domain will be the set of natural numbers. {1, 2, 3, 4, . . .}
    ♦ Then it will be an infinite sequence.
• In some cases, the domain will be a subset of natural numbers. {1, 2, 3, 4, . . . k}
    ♦ Then it will be a finite sequence.
• The range will be the terms of the sequence.

---

• There are some sequences which do not have an algebraic formula for the n^th term.
• For example, consider the sequence 2, 3, 5, 7, 11, 13, 17, . . .
• It is the sequence of prime numbers. There is no algebraic formula to obtain the terms of this sequence.
• In such cases, we describe the sequence by a verbal description. This verbal description is called the rule for generating the terms of the sequence.

---

Series

Basics about series can be written in steps:
1. Let a₁, a₂, a₃, a₄, . . . a_n be a given sequence.
• Then the expression a₁ + a₂ + a₃ + a₄ + . . . a_n + . . . is called the series associated with the given sequence.
2. Note that the expression a₁ + a₂ + a₃ + a₄ + . . . a_n + . . . is the series.
• We must not consider the actual sum as the series.
• The actual sum is referred to as sum of the series.
3. We can write about finite and infinite series:
• If the given sequence is finite, then the associated series is a finite series.
• If the given sequence is infinite, then the associated series is an infinite series.
4. Since the series involves a summation, we can use the sigma notation. We have seen the details about sigma notation in a previous chapter. [see section 8.1]
• So the series can be abbreviated as: $\sum\limits_{k\,=\,1}^{k\,=\,n}{a_k}$
• 'k=1' at the bottom of '∑' indicates that, the value of k starts from 1. In other words, the value of k in the first term is 1.
• 'k=n' at the top of '∑' indicates that, the value of k ends at n. In other words, the value of k in the last term is n.
• So the values of k are: 1, 2, 3, . . . n
• It is known as the summation from k = 1 to k = n.
• Let us write an example:
The series associated with the sequence of odd natural numbers can be written in two forms:
(i) 1 + 3 + 5 + 7 + . . .
(ii) $\sum\limits_{k\,=\,1}^{k\,=\,\infty}{2k-1}$

---

In the next section we will see some solved examples on sequences and series.

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Chapter 8.5 - Miscellaneous Examples on Chapter 8

In the previous section, we completed a discussion on General and Middle terms in binomial expansions. In this section, we will see some miscellaneous examples.

Solved example 8.10
Find the term independent of x in the expansion of $\left({\frac{3}{2}}x^2~-~\frac{1}{3x} \right)^6$.
Solution:
1. Assume that the term independent of x occurs in the (r+1)^th term.
• We know that, the (r+1)^th term of the binomial expansion (a-b)ⁿ is given by:
ⁿC_r (-1)^r a^n-r b^r
• In our present case, n = 6,  $a={\frac{3}{2}}x^2,~~b=\frac{1}{3x}$
2. So we can write:
(r+1)^th term of the expansion of $\left({\frac{3}{2}}x^2~-~\frac{1}{3x} \right)^6~=~{}^{6} {\rm{C}}_r\;(-1)^r \; \left({\frac{3}{2}}x^2 \right)^{6-r} \;\left(\frac{1}{3x} \right)^r$
This can be simplified as:
${}^{6} {\rm{C}}_r\;(-1)^r \; \left({\frac{3}{2}}\right)^{6-r}\; \left(x^2 \right)^{6-r} \;\left(\frac{1}{3^r x^r} \right)~=~{}^{6} {\rm{C}}_r\;(-1)^r \; \left({\frac{3}{2}}\right)^{6-r}\; \left(x^{12-2r} \right)\;\left(\frac{1}{3^r x^r} \right)$
3. Let us compare the indices of x:
    ♦ The index of x in the numerator is 12-2r
    ♦ The index of x in the denominator is r
• These two indices must be equal. So we get: 12 - 2r = r
⇒ 12 = 3r, which gives: r = 4
4. So we can write:
The term with r = 4, will be independent of x. That means, the fifth term will be independent of x.
• From (2), we get:
(4+1)^th term = ${}^{6} {\rm{C}}_4\;(-1)^4 \; \left({\frac{3}{2}}x^2 \right)^{6-4} \;\left(\frac{1}{3x} \right)^4~=~{}^{6} {\rm{C}}_4\; × \;1 \; × \; \left({\frac{3}{2}}x^2 \right)^{2} \;\left(\frac{1}{3^4 x^4} \right)$
$~=~15 × \; \left({\frac{3^2}{2^2}}\right)\; × x^4 \; × \;\left(\frac{1}{3^4 x^4} \right)~=~\frac{15}{36}~=~\frac{5}{12}$

Solved example 8.11
If the coefficients of a^r-1, a^r and a^r+1 in the expansion of (1 + a)ⁿ are in arithmetic progression, prove that n² – n(4r + 1) + 4r² – 2 = 0.
Solution:
1. We know that, the (r+1)^th term of the binomial expansion (a+b)ⁿ is given by:
ⁿC_r a^n-r b^r
• In our present case, n = n, a = 1 and b = a
2. So we can write:
(r+1)^th term of the expansion of (1 + a)ⁿ is: ⁿC_r 1^n-r a^r
= ⁿC_r a^r
3. It follows that:
• r^th term of the expansion of (1 + a)ⁿ will be: ⁿC_r-1 a^r-1
• (r+2)^th term of the expansion of (1 + a)ⁿ will be: ⁿC_r+1 a^r+1
4. So the three coefficients in arithmetic progression are:
ⁿC_r-1, ⁿC_r, ⁿC_r+1
• Since they are in arithmetic progression, the common difference will be the same. We can write:
ⁿC_r  - ⁿC_r-1 = ⁿC_r+1  - ⁿC_r 
^{⇒ n}C_r-1 + ⁿC_r+1 = 2 × ⁿC_r
• This can be simplified as:
$\begin{array}{ll}
{}&{}
&{\frac{n!}{(r-1)!(n-r+1)!}~+~\frac{n!}{(r+1)!(n-r-1)!}}& {}={}
&{2 × \frac{n!}{r!(n-r)!}}& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{}&{}\Rightarrow{}
&{\frac{n!}{(r-1)!(n-r+1)(n-r)(n-r-1)!}~+~\frac{n!}{(r+1)r(r-1)!(n-r-1)!}}& {}={}
&2 × \frac{n!}{r!(n-r)!}& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{}&{}\Rightarrow{}
&{\frac{n!r(r+1)~+~n!(n-r+1)(n-r)}{(r-1)!(n-r+1)(n-r)(n-r-1)!r (r+1)}}& {}={}
&2 × \frac{n!}{r!(n-r)!}& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{}&{}\Rightarrow{}
&{\frac{n!r(r+1)~+~n!(n-r+1)(n-r)}{[r(r+1)(r-1)!][(n-r+1)(n-r)(n-r-1)!]}}& {}={}
&2 × \frac{n!}{r!(n-r)!}& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{}&{}\Rightarrow{}
&{\frac{n!r(r+1)~+~n!(n-r+1)(n-r)}{[(r+1)!][(n-r+1)(n-r)!]}}& {}={}
&2 × \frac{n!}{r!(n-r)!}& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{}&{}
&\color {green} {\text{[n! and (n-r)! can be cancelled from both sides]}}& {}&{}& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{}&{}\Rightarrow{}
&{\frac{r(r+1)~+~(n-r+1)(n-r)}{[(r+1)!][(n-r+1)]}}& {}={}
&2 × \frac{1}{r!}& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{}&{}\Rightarrow{}
&{\frac{r(r+1)~+~(n-r+1)(n-r)}{[(r+1)r!][(n-r+1)]}}& {}={}
&2 × \frac{1}{r!}& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{}&{}
&\color {green} {\text{[r! can be cancelled from both sides]}}& {}&{}& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{}&{}\Rightarrow{}
&{\frac{r(r+1)~+~(n-r+1)(n-r)}{[(r+1)][(n-r+1)]}}& {}={}
&2 × 1& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{}&{}\Rightarrow{}
&{r^2+r+n^2-nr-nr+r^2+n-r}& {}={}
&2 × [nr-r^2+r+n-r+1]& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{}&{}\Rightarrow{}
&{2r^2+n^2-2nr+n}& {}={}
&2nr-2r^2+2n+2& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{}&{}\Rightarrow{}
&{4r^2+n^2-4nr-n-2}& {}={}
&0& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{}&{}\Rightarrow{}
&{n^2-n(4r+1)+4r^2-2}& {}={}
&0& {}
&{}& {}
&{}& {}&{}& {} &{} \\

\end{array}$

Solved example 8.12
Show that the coefficient of the middle term in the expansion of (1 + x)²ⁿ is equal to the sum of the coefficients of two middle terms in the expansion of (1 + x)^2n-1.
Solution:
1. Consider (1 + x)²ⁿ. The index will be always even. So there will always be an odd number of terms. As a result, there will be a single middle term.
• The position of that single middle term is: $\frac{n}{2}~+~1$
• In our present case, n = 2n.
• So position of the middle term is $\frac{2n}{2}~+~1~=~(n+1)$
• We know that, the (r+1)^th term of the binomial expansion (a+b)ⁿ is given by:
ⁿC_r a^n-r b^r
• In our present case, n = 2n, a = 1 and b = x
• So we can write:
(r+1)^th term of the expansion of (1 + x)²ⁿ is: ²ⁿC_r 1^2n-r a^r
= ²ⁿC_r a^r
• Based on this, we can write:
Middle term, which is the (n+1)^th term will be: ²ⁿC_n aⁿ
• Coefficient of this middle term is ²ⁿC_n
• This can be simplified as:
$\frac{(2n)!}{n!(2n-n)!}~=~\frac{2n(2n-1)!}{n!n!}~=~\frac{2n(2n-1)!}{n!n(n-1)!}~=~\frac{2(2n-1)!}{n!(n-1)!}$
2. Consider (1 + x)^2n-1. The index will be always odd. So there will always be an even number of terms. As a result, there will be a two middle terms.
• The positions of those middle terms are: $\frac{n+1}{2}~\text{and}~\frac{n+1}{2}~+~1$
• In our present case, n = 2n-1.
• So position of the middle terms are $\frac{(2n-1)+1}{2}~\text{and}~\frac{(2n-1)+1}{2}~+~1$
• So the positions are: n and (n+1)
• We know that, the (r+1)^th term of the binomial expansion (a+b)ⁿ is given by:
ⁿC_r a^n-r b^r
• In our present case, n = 2n-1, a = 1 and b = x
• So we can write:
(r+1)^th term of the expansion of (1 + x)^2n-1 is: ^2n-1C_r 1^2n-1-r a^r
= ^2n-1C_r a^r
• Based on this, we can write:
   ♦ The second middle term, which is the (n+1)^th term will be: ^2n-1C_n aⁿ
   ♦ The first middle term, which is the n^th term will be: ^2n-1C_n-1 a^n-1
• Sum of the coefficients of these two middle terms = ^2n-1C_n + ^2n-1C_n-1
This can be simplified as:
$\frac{(2n-1)!}{n!(2n-1-n)!}~+~\frac{(2n-1)!}{(n-1)!(2n-1-n+1)!}$
= $\frac{(2n-1)!}{n!(n-1)!}~+~\frac{(2n-1)!}{(n-1)!n!}$
= $\frac{2(2n-1)!}{n!(n-1)!}$
3. Comparing the results in (1), and (2), we see that:
Result in (2) is same as the result in (1)

Solved example 8.13
Find the coefficient of a⁴ in the product (1 + 2a)⁴ (2 – a)⁵ using binomial
theorem.
Solution:
1. Let us expand each of the factors of the given product. We get:
• First factor $(1+2a)^4$
$\begin{array}{ll}
{}={}&{}^4 {\rm{C}}_0  × 1^{4-0} × (2a)^0
&{}+{}& {}^4 {\rm{C}}_1 × 1^{4-1} × (2a)^1
&{}+{}& {}^4 {\rm{C}}_2  × 1^{4-2} × (2a)^2
&{}+{}& {}^4 {\rm{C}}_3 × 1^{4-3} × (2a)^3
&{}+{}& {}^4 {\rm{C}}_4 × 1^{4-4} × (2a)^4 \\

{}={}&{}^4 {\rm{C}}_0  × 1 × 1
&{}+{}& {}^4 {\rm{C}}_1 × 1 × (2a)
&{}+{}& {}^4 {\rm{C}}_2 × 1 × 4a^2
&{}+{}& {}^4 {\rm{C}}_3 × 1 × 8a^3
&{}+{}& {}^4 {\rm{C}}_4 × 1 × 16a^4 \\

{}={}&1
&{}+{}&8a
&{}+{}& 24a^2
&{}+{}& 32a^3
&{}+{}& 16a^4 \\
\end{array}$

• Second factor $(2-a)^5$
$\begin{array}{ll}
{}={}&{}^5 {\rm{C}}_0  × (-1)^0 × 2^{5-0} × a^0
&{}+{}& {}^5 {\rm{C}}_1  × (-1)^1 × 2^{5-1} × a^1
&{}+{}& {}^5 {\rm{C}}_2  × (-1)^2 × 2^{5-2} × a^2
&{}+{}& {}^5 {\rm{C}}_3  × (-1)^3 × 2^{5-3} × a^3
&{}+{}& {}^5 {\rm{C}}_4  × (-1)^4 × 2^{5-4} × a^4
&{}+{}& {}^5 {\rm{C}}_5  × (-1)^5 × 2^{5-5} × a^5 \\

{}={}&{}^5 {\rm{C}}_0  × 2^5 × 1
&{}-{}& {}^5 {\rm{C}}_1 × 2^4 × a
&{}+{}& {}^5 {\rm{C}}_2 × 2^3 × a^2
&{}-{}& {}^5 {\rm{C}}_3 × 2^2 × a^3
&{}+{}& {}^5 {\rm{C}}_4 × 2^1 × a^4
&{}-{}& {}^5 {\rm{C}}_5 × 2^0 × a^5 \\

{}={}&1  × 2^5 × 1
&{}-{}& 5 × 2^4 × a
&{}+{}& 10 × 2^3 × a^2
&{}-{}& 10 × 2^2 × a^3
&{}+{}& 5 × 2^1 × a^4
&{}-{}& 1 × 2^0 × a^5 \\

{}={}&32
&{}-{}& 80a
&{}+{}& 80a^2
&{}-{}& 40a^3
&{}+{}& 10a^4
&{}-{}& a^5 \\
\end{array}$

2. So we can write:
(1 + 2a)⁴  × (2 – a)⁵ =
$\left(1+8a+24a^2+32a^3+16a^4 \right) × \left(32-80a+80a^2-40a^3+10a^4-a^5 \right)$
3. There is no need to do the actual multiplication. There is an easier method to find just the required terms. It can be explained in 4 steps:
(i) Pick each term from the first factor.
(ii) Try to pair it with each term of the second factor.
(iii) All the pairs which give a⁴ are to be selected.
(iv) All the pairs which do not give a⁴ are to be discarded.
4. Thus we get:
$[1 × 10a^4] + [8a × -40a^3]+[24a^2 × 80a^2]+[32a^3 × -80a]+[16a^4 × 32]$
= $[1 × 10a^4] - [8a × 40a^3]+[24a^2 × 80a^2]-[32a^3 × 80a]+[16a^4 × 32]$
= $10a^4-320a^4+1920a^4-2560a^4+512a^4$
= $-438a^4$
• So the coefficient of a⁴ in the given product is -438

Solved example 8.14
Find the r^th term from the end in the expansion of (x + a)ⁿ.
Solution:
1. We know that, the (r+1)^th term of the binomial expansion (a+b)ⁿ is given by:
ⁿC_r a^n-r b^r
• In our present case, a = x and b = a.
• So we get:
(r+1)^th term of the binomial expansion (x+a)ⁿ is: ⁿC_r x^n-r a^r
2. In the expansion , there will be (n+1) terms. That means, the (n+1)^th term will be the last term.
• Using the result in (1), we can write:
The last term, which is the (n+1)^th term will be ⁿC_n x^n-n aⁿ 
3. But the last term, is the first term from the end. So we can write:
First term from the end will be ⁿC_n x^n-n aⁿ 
4. Now, the n^th term will be the second term from the end.
• Using the result in (1), we can write:
The second term from the end, which is the n^th term will be ⁿC_n-1 x^n-(n-1) a^n-1 
5. similarly, the (n-1)^th term will be the third term from the end.
• Using the result in (1), we can write:
The third term from the end, which is the (n-1)^th term will be ⁿC_n-2 x^n-(n-2) a^n-2 
6. similarly, the (n-2)^th term will be the fourth term from the end.
• Using the result in (1), we can write:
The fourth term from the end, which is the (n-2)^th term will be ⁿC_n-3 x^n-(n-3) a^n-3 
7. Thus we see a pattern. Based on that pattern, we can write:
The r^th term from the end will be ⁿC_n-(r-1) x^n-[n-(r-1)] a^n-(r-1) 
• Simplifying this, we get: ⁿC_n-r+1 x^n-n+r-1 a^n-r+1 
= ⁿC_n-r+1 x^r-1 a^n-r+1

Solved example 8.15
Find the term independent of x in the expansion of $\left(\sqrt[3]{x}~+~\frac{1}{2 \sqrt[3]{x}} \right)^{18},~x>0$.
Solution:
1. We know that, the (r+1)^th term of the binomial expansion (a+b)ⁿ is given by:
ⁿC_r a^n-r b^r
• In our present case, $a~=~\sqrt[3]{x}~~\text{and}~~b~=~\frac{1}{2 \sqrt[3]{x}}$.
• So we get:
(r+1)^th term of the given binomial expansion is:
${}^{n} {\rm{C}}_{r}~\left(\sqrt[3]{x} \right)^{n-r}~\left(\frac{1}{2 \sqrt[3]{x}} \right)^r$
2. This can be simplified as:
${}^{n} {\rm{C}}_{r}~\left(x^{\frac{1}{3}} \right)^{n-r}~\left(\frac{1}{2} \right)^r~\left(\frac{1}{x^{\frac{1}{3}}} \right)^r$

= ${}^{n} {\rm{C}}_{r}~\left(x \right)^{^{\frac{n-r}{3}}}~\left(\frac{1}{2} \right)^r~\left(\frac{1}{x^{\frac{r}{3}}} \right)$
3. Now we compare the powers:
   ♦ Power of x in the numerator
   ♦ must be equal to
   ♦ Power of x in the denominator
• Thus we get: $\frac{n-r}{3}~=~\frac{r}{3}$
⇒ n-r = r
⇒ $r~=~\frac{n}{2}~=~\frac{18}{2}~=~9$
4. So we can write:
The term with r = 9, will be independent of x.
That means, the 10^th term will be independent of x.
5. From the result in (2), we get:
$T_{9+1}~=~{}^{18} {\rm{C}}_{9}~\left(x \right)^{^{\frac{18-9}{3}}}~\left(\frac{1}{2} \right)^9~\left(\frac{1}{x^{\frac{9}{3}}} \right)~=~{}^{18} {\rm{C}}_{9}~\left(x \right)^{^{\frac{9}{3}}}~\left(\frac{1}{2} \right)^9~\left(\frac{1}{x^{\frac{9}{3}}} \right)~=~{}^{18} {\rm{C}}_{9}~\left(\frac{1}{2} \right)^9~=~{}^{18} {\rm{C}}_{9}~\left(\frac{1}{2^9} \right)$

Solved example 8.16
The sum of the coefficients of the first three terms in the expansion of $\left(x-\frac{3}{x^2} \right)^m,~x \ne 0$, m being a natural number, is 559. Find the term of the expansion containing x³ .
Solution:
1. We know that, the (r+1)^th term of the binomial expansion (a-b)ⁿ is given by:
ⁿC_r (-1)^r a^n-r b^r
• In our present case, $n~=~m,~a~=~x~~\text{and}~~b~=~\frac{3}{x^2}$.
• So we get:
(r+1)^th term of the given binomial expansion is:
${}^{m} {\rm{C}}_{r}~(-1)^r~\left(x \right)^{m-r}~\left(\frac{3}{x^2} \right)^r$
2. Now we can write the terms:
◼ For the first term, we put r = 0. So we get:
$T_{0+1}~=~{}^{m} {\rm{C}}_{0}~(-1)^0~\left(x \right)^{m-0}~\left(\frac{3}{x^2} \right)^0~=~{}^{m} {\rm{C}}_{0}~ × 1 × x^m × 1~=~\left[{}^{m} {\rm{C}}_{0}\right]\;\left[x^m \right]$
◼ For the second term, we put r = 1. So we get:
$T_{1+1}~=~{}^{m} {\rm{C}}_{1}~ (-1)^1\left(x \right)^{m-1}~\left(\frac{3}{x^2} \right)^1~=~{}^{m} {\rm{C}}_{1}~ × -1 × x^{m-1}~\left(\frac{3^1}{x^{2 × 1}} \right)~=~-\left[{}^{m} {\rm{C}}_{1}\;3^1 \right]\;\left[\frac{x^{m-1}}{x^{2 × 1}}\right]~=~-\left[{}^{m} {\rm{C}}_{1} × 3 \right]\;\left[\frac{x^{m-1}}{x^{2}}\right]$
◼ For the third term, we put r = 2. So we get:
$T_{2+1}~=~{}^{m} {\rm{C}}_{2}~(-1)^2~\left(x \right)^{m-2}~\left(\frac{3}{x^2} \right)^2~=~{}^{m} {\rm{C}}_{2}~ × 1 × ~x^{m-2}~\left(\frac{3^2}{x^{2 × 2}} \right)~=~\left[{}^{m} {\rm{C}}_{2}\;3^2 \right]\;\left[\frac{x^{m-2}}{x^{2 × 2}}\right]~=~\left[{}^{m} {\rm{C}}_{2} × 3^2 \right]\;\left[\frac{x^{m-2}}{x^{4}}\right]$
3. So the coefficients are: $\left[{}^{m} {\rm{C}}_{0}\right],~-\left[{}^{m} {\rm{C}}_{1} × 3 \right],~~\text{and}~~\left[{}^{m} {\rm{C}}_{2} × 3^2 \right]$
We can write:
Sum of the coefficients = $\left[{}^{m} {\rm{C}}_{0}\right]~-~\left[{}^{m} {\rm{C}}_{1} × 3 \right]~+~\left[{}^{m} {\rm{C}}_{2} × 3^2 \right]~=~559$
4. This can be simplified as:
$\frac{m!}{0!(m-0)!}~-~\frac{3 × m!}{1!(m-1)!}~+~\frac{9 × m!}{2!(m-2)!}~=~559$
$\Rightarrow~\frac{m!}{0!\;m!}~-~\frac{3m(m-1)!}{1!(m-1)!}~+~\frac{9m(m-1)(m-2)!}{2!(m-2)!}~=~559$
$\Rightarrow~1~-~\frac{3m}{1}~+~\frac{9m(m-1)}{2}~=~559$
$\Rightarrow~2~-~6m~+~9m(m-1)~=~559 × 2$
$\Rightarrow~2~-~6m~+~9m^2-9m~=~1118$
$\Rightarrow~9m^2 - 15m~=~1116$
• Solving this quadratic equation, we get: m = 12
5. Based on the result in (1), we can write:
For the power of x to be 3, [m-r] - 2r should be 3
So we get: [12-r] - 2r = 3
12 - 3r = 3
r = 3
6. Thus we can write:
In the term with r = 3, the power of x will be 3
• We get:
$T_{r+1}~=~T_{3+1}~=~{}^{12} {\rm{C}}_{3}~(-1)^3~\left(x \right)^{12-3}~\left(\frac{3}{x^2} \right)^r~=~-5940x^3$

Solved example 8.17
If the coefficients of (r – 5)^th and (2r – 1)^th terms in the expansion of (1 + x)³⁴ are equal, find r.
Solution:
1. We know that, the (r+1)^th term of the binomial expansion (a+b)ⁿ is given by:
ⁿC_r a^n-r b^r
• In our present case, a = 1, b = x and n = 34.
• So we get:
(r+1)^th term of the given binomial expansion is:
$T_{r+1}~=~{}^{34} {\rm{C}}_{r}~1^{34-r}~x^r~=~{}^{34} {\rm{C}}_{r}~x^r$
2. Based on this, we can write:
r^th term of the given binomial expansion is:
$T_{r}~=~{}^{34} {\rm{C}}_{r-1}~x^{r-1}$
3. So the (r – 5)^th term will be:
$T_{r-5}~=~{}^{34} {\rm{C}}_{(r-5)-1}~x^{(r-5)-1}~=~{}^{34} {\rm{C}}_{r-6}~x^{r-6}$
4. Similarly, the (2r – 1)^th term will be:
$T_{2r-1}~=~{}^{34} {\rm{C}}_{(2r-1)-1}~x^{(2r-1)-1}~=~{}^{34} {\rm{C}}_{2r-2}~x^{2r-2}$
5. Given that, the coefficients are equal. So we can write:
${}^{34} {\rm{C}}_{r-6}~=~{}^{34} {\rm{C}}_{2r-2}$
6. Now we apply a property that we saw in the previous chapter:
If ${}^{n} {\rm{C}}_{a}~=~{}^{n} {\rm{C}}_{b}$, then a = b or n = a+b
[See result 6 in section 7.6]
• We can write:
(i) r - 6 = 2r-2  OR
(ii) r - 6 + 2r - 2 = 34
   ♦ From (i), we get: r = -4
   ♦ From (ii), we get: r = 14
• r must be a +ve integer. So r = 14

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The link below gives some more miscellaneous examples.

Miscellaneous Exercise on chapter 8

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In the next chapter we will see sequences and series.

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Chapter 8.4 - Solved Examples on General and Middle Terms

In the previous section, we saw General and Middle terms in binomial expansions. We saw a solved example also. In this section, we will see a few more solved examples.

Solved example 8.6
Show that the middle term in the expansion of $(1+x)^{2n}$ is $\frac{1 × 3 × 5~.~.~.~(2n-1)}{n!} 2n x^n$, where n is a positive integer.
Solution:
1. Given that, n is a +ve integer. So 2n will be an even number.
2. Since the index is 2n, there will be an odd number (2n+1) of terms.
3. We have seen that, if the number of terms is odd, there will be an unique middle term.
• It's position is given by: $\frac{2n}{2}+1~=~(n+1)$
4. Now consider the expansion of $(1+x)^{2n}$
• We want to find the middle term, which is the (n+1)^th term.
• We know that, the (r+1)^th term of the binomial expansion (a+b)ⁿ is given by: ⁿC_r a^n-r b^r
5. In our present case, a = 1, b = x and n = 2n.
So the middle term, which is the (n+1)^th term will be:
${}^{2n} {\rm{C}}_n\;1^{2n-n}\;x^n$
$\begin{array}{ll}
{}={}&{}^{2n} {\rm{C}}_n\; × 1 × x^n&{}& {} &{} \\
{}={}&\frac{2n! × x^n}{n!(2n-n)!}&{}& {} &{} \\
{}={}&\frac{2n! × x^n}{n! n!}&{}& {} &{} \\
{}={}&\frac{2n(2n-1)(2n-2)~.~.~.~4 × 3 × 2 × 1 × x^n}{n! n!}&{}& {} &{} \\
{}={}&\frac{1 × 2 × 3 × 4~.~.~.~(2n-2)(2n-1)2n× x^n}{n! n!}&{}& {} &{} \\
{}={}&\frac{1 × 2 × 3 × 4~.~.~.~(2n-1)2n× x^n}{n! n!}&{}& {} &{} \\
{}={}&\frac{[1 × 3 × 5 × ~.~.~.~ ×(2n-1)] [2 × 4 × 6 × ~.~.~.~ × 2n]× x^n}{n! n!}&{}& {} &{} \\
{}={}&\frac{[1 × 3 × 5 × ~.~.~.~ ×(2n-1)] [(2 × 1) × (2 × 2) × (2 × 3) × ~.~.~.~ × (2 × n)]× x^n}{n! n!}&{}& {} &{} \\
{}={}&\frac{[1 × 3 × 5 × ~.~.~.~ ×(2n-1)] 2^n[1 × 2 × 3 × ~.~.~.~ × n]× x^n}{n! n!}&{}& {} &{} \\
{}={}&\frac{[1 × 3 × 5 × ~.~.~.~ ×(2n-1)] 2^n[n!]× x^n}{n! n!}&{}& {} &{} \\
{}={}&\frac{[1 × 3 × 5 × ~.~.~.~ ×(2n-1)] 2^n× x^n}{n!}&{}& {} &{} \\
\end{array}$

Solved example 8.7
Find the coefficient of x⁶y³ in the expansion of (x+2y)⁹
Solution:
1. Assume that x⁶y³ occurs in the (r+1)^th term.
• We know that, the (r+1)^th term of the binomial expansion (a+b)ⁿ is given by: ⁿC_r a^n-r b^r
• In our present case, n = 9, a = x and b = 2y
2. So we can write:
(r+1)^th term of the expansion of (x+2y)⁹ = ⁹C_r x^9-r (2y)^r
= ⁹C_r x^9-r 2^r y^r = [⁹C_r  × 2^r][x^9-r  × y^r]
• Thus we get:
   ♦ Constant part (coefficient) of the (r+1)^th term  = [⁹C_r  × 2^r]
   ♦ Variable part of the (r+1)^th term = [x^9-r  × y^r]
3. But given that, the variable part is x⁶y³
Comparing this with [x^9-r  × y^r], we get: r = 3
4. So the coefficient = [⁹C_r  × 2^r] = [⁹C₃  × 2³] = 672

Solved example 8.8
The second, third and fourth terms in the binomial expansion (x + a)ⁿ are
240, 720 and 1080, respectively. Find x, a and n.
Solution:
1. We know that, the (r+1)^th term of the binomial expansion (a+b)ⁿ is given by: ⁿC_r a^n-r b^r
• In our present case, a = x and b = a
2. For the second term, r = 1
So we get: T₂ = ⁿC₁ x^n-1 a¹ = 240  
3. For the third term, r = 2
So we get: T₃ = ⁿC₂ x^n-2 a² = 720  
4. For the fourth term, r = 3
So we get: T₄ = ⁿC₃ x^n-3 a³ = 1080
5. Dividing (3) by (2), we get:
$\begin{array}{ll}
{\frac{T_3}{T_2}}&{}={}
&\frac{{}^{n} {\rm{C}}_2 × x^{n-2} × a^2}{{}^{n} {\rm{C}}_1 × x^{n-1} × a^1}& {}={}
&\frac{720}{240}& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{}&{}={}
&{\frac{{}^{n} {\rm{C}}_2 × x^n × x^{-2} × a^2}{{}^{n} {\rm{C}}_1 × x^n × x^{-1} × a^1}}& {}={}
&3& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{}&{}={}
&{\frac{{}^{n} {\rm{C}}_2 × x^{-1} × a}{{}^{n} {\rm{C}}_1 }}& {}={}
&3& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{}&{}={}
&{\frac{\frac{n!}{2!(n-2)!} × x^{-1} × a}{\frac{n!}{1!(n-1)!} }}& {}={}
&3& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{}&{}={}
&{\frac{1!(n-1)! × x^{-1} × a}{2!(n-2)! }}& {}={}
&3& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{}&{}={}
&{\frac{(n-1)(n-2)! × x^{-1} × a}{2 × (n-2)! }}& {}={}
&3& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{}&{}={}
&{\frac{(n-1) × x^{-1} × a}{2}}& {}={}
&3& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{}&{}={}
&{\frac{(n-1)}{2}~ × ~\frac{a}{x}}& {}={}
&3& {}
&{}& {}
&{}& {}&{}& {} &{} \\

\end{array}$

$\Rightarrow~\frac{a}{x}~=~\frac{6}{(n-1)}$ 
6. Dividing (4) by (3), we get:
$\begin{array}{ll}
{\frac{T_4}{T_3}}&{}={}
&\frac{{}^{n} {\rm{C}}_3 × x^{n-3} × a^3}{{}^{n} {\rm{C}}_2 × x^{n-2} × a^2}& {}={}
&\frac{1080}{720}& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{}&{}={}
&{\frac{{}^{n} {\rm{C}}_3 × x^n × x^{-3} × a^3}{{}^{n} {\rm{C}}_2 × x^n × x^{-2} × a^2}}& {}={}
&1.5& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{}&{}={}
&{\frac{{}^{n} {\rm{C}}_3 × x^{-1} × a}{{}^{n} {\rm{C}}_2 }}& {}={}
&1.5& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{}&{}={}
&{\frac{\frac{n!}{3!(n-3)!} × x^{-1} × a}{\frac{n!}{2!(n-2)!} }}& {}={}
&1.5& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{}&{}={}
&{\frac{2!(n-2)! × x^{-1} × a}{3!(n-3)! }}& {}={}
&1.5& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{}&{}={}
&{\frac{2! × (n-2)(n-3)! × x^{-1} × a}{3 × 2! × (n-3)! }}& {}={}
&1.5& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{}&{}={}
&{\frac{(n-2) × x^{-1} × a}{3}}& {}={}
&1.5& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{}&{}={}
&{\frac{(n-2)}{3}~ × ~\frac{a}{x}}& {}={}
&1.5& {}
&{}& {}
&{}& {}&{}& {} &{} \\

\end{array}$
$\Rightarrow~\frac{a}{x}~=~\frac{4.5}{(n-2)}$
7. Equating the results in (5) and (6), we get:
$\frac{a}{x}~=~\frac{6}{(n-1)}~=~\frac{4.5}{n-2}$
$\Rightarrow~6n-12~=~4.5n-4.5$ 
$\Rightarrow~ 1.5n~=~7.5$
$\Rightarrow~ n ~=~5$
8. Substituting the value of n in (5), we get:
$\frac{a}{x}~=~\frac{6}{5-1}~=~\frac{6}{4}~=~1.5$
$\Rightarrow~a~=~1.5x$
9. Substituting the value of n in (2), we get:
⁵C₁ x^5-1 a¹ = 240  
⇒ 5 × x⁴× a = 240
• Substituting for a using (8), we get:
5 × x⁴ × 1.5x = 240
⇒ 7.5x⁵ = 240
⇒ x = 2
10. Substituting for x in (8), we get:
a = 1.5 × 2 = 3

Solved example 8.9
The coefficients of three consecutive terms in the expansion of (1 + a)ⁿ are in the ratio 1: 7 : 42. Find n.
Solution:
1. Let the three consecutive terms be (r+1)^th, (r+2)^th and (r+3)^th
2. The general term is given by: T_r+1 = ⁿC_r a^n-r b^r
• In our present case, a = 1 and b = a
3. The first of the three consecutive terms will be: ⁿC_r+1 1^n-1 a^r+1 = ⁿC_r+1 a^r+1
So the coefficient of this first term will be ⁿC_r+1
4. The second of the three consecutive terms will be: ⁿC_r+2 1^n-2 a^r+2 = ⁿC_r+2 a^r+2
So the coefficient of this second term will be ⁿC_r+2
5. The third of the three consecutive terms will be: ⁿC_r+3 1^n-3 a^r+3 = ⁿC_r+3 a^r+3
So the coefficient of this third term will be ⁿC_r+3
6. We are given the ratio between the coefficients. So we can write:
ⁿC_r+1 : ⁿC_r+2 : ⁿC_r+3 = 1 : 7 : 42
7. Consider the ratio of second coefficient to first coefficient. We get:
$\frac{{}^{n} {\rm{C}}_{r+2}}{{}^{n} {\rm{C}}_{r+1}}~=~7$

$\Rightarrow~\frac{\frac{n!}{(r+2)![n-(r+2)]!}}{\frac{n!}{(r+1)![n-(r+1)]!}}~=~7$

$\Rightarrow~\frac{(r+1)![n-(r+1)]!}{(r+2)![n-(r+2)]!}~=~7$

$\Rightarrow~\frac{(r+1)![n-(r+1)]!}{(r+2)(r+1)![n-(r+2)]!}~=~7$

$\Rightarrow~\frac{[n-(r+1)]!}{(r+2)[n-(r+2)]!}~=~7$

$\Rightarrow~\frac{[n-r-1]!}{(r+2)[n-r-2]!}~=~7$

$\Rightarrow~\frac{[n-r-1][n-r-2]!}{(r+2)[n-r-2]!}~=~7$

$\Rightarrow~\frac{[n-r-1]}{(r+2)}~=~7$

$\Rightarrow~n-r-1~=~7r+14$

$\Rightarrow~n-8r~=~15$

8. Consider the ratio of third coefficient to second coefficient. We get:
$\frac{{}^{n} {\rm{C}}_{r+3}}{{}^{n} {\rm{C}}_{r+2}}~=~\frac{42}{7}~=~6$

$\Rightarrow~\frac{\frac{n!}{(r+3)![n-(r+3)]!}}{\frac{n!}{(r+2)![n-(r+2)]!}}~=~7$

$\Rightarrow~\frac{(r+2)![n-(r+2)]!}{(r+3)![n-(r+3)]!}~=~6$

$\Rightarrow~\frac{(r+2)![n-(r+2)]!}{(r+3)(r+2)![n-(r+3)]!}~=~6$

$\Rightarrow~\frac{[n-(r+2)]!}{(r+3)[n-(r+3)]!}~=~6$

$\Rightarrow~\frac{[n-r-2]!}{(r+3)[n-r-3]!}~=~6$

$\Rightarrow~\frac{[n-r-2][n-r-3]!}{(r+3)[n-r-3]!}~=~6$

$\Rightarrow~\frac{[n-r-2]}{(r+3)}~=~6$

$\Rightarrow~n-r-2~=~6r+18$

$\Rightarrow~n-7r~=~20$

9. So we have two equations:
(i) From (7), we have: n-8r = 15
(ii) From (8), we have: n-7r = 20
Solving these two equations, we get: r = 5 and n = 55

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The link below gives some more solved examples.

Exercise 8.2

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In the next section we will see some miscellaneous examples.

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Chapter 8.3 - General and Middle Terms

In the previous section, we saw some solved examples on how to expand binomials. In this section, we will see General and Middle terms in binomial expansions.

Some basics about general term can be written in 5 steps:
1. In the binomial expansion for (a+b)ⁿ, let us denote the position of any term by r. Then:
   ♦ For the first term, r = 1   
   ♦ For the second term, r = 2   
   ♦ For the third term, r = 3   
   ♦ For the fourth term, r = 4
   ♦ so on . . .
2. Now let us consider each term of the expansion:
• When r = 1, the term is ⁿC₀ a^n-0 b⁰
   ♦ We see that:
         ✰ The subscript of C is (r-1)
         ✰ The power of a is [n-(r-1)]
         ✰ The power of b is (r-1) 
• When r = 2, the term is ⁿC₁ a^n-1 b¹
   ♦ We see that:
         ✰ The subscript of C is (r-1)
         ✰ The power of a is [n-(r-1)]
         ✰ The power of b is (r-1) 
• When r = 3, the term is ⁿC₂ a^n-2 b²
   ♦ We see that:
         ✰ The subscript of C is (r-1)
         ✰ The power of a is [n-(r-1)]
         ✰ The power of b is (r-1)
◼ So we see a definite pattern in the subscript of C, power of a and power of b.
3. We see that (r-1) occurs frequently.
• This is inconvenient. We want r instead of (r-1).
• For that, we consider the (r+1)^th term instead of the r^th term.
4. Let us see the new pattern:
• When r = 1, we get: (r+1) = 2
So we consider the second term, which is: ⁿC₁ a^n-1 b¹
   ♦ We see that:
         ✰ The subscript of C is r
         ✰ The power of a is [n-r]
         ✰ The power of b is r   
• When r = 2, we get: (r+1) = 3
So we consider the third term, which is: ⁿC₂ a^n-2 b²
   ♦ We see that:
         ✰ The subscript of C is r
         ✰ The power of a is [n-r]
         ✰ The power of b is r   
• When r = 1, we get: (r+1) = 2
So we consider the second term, which is: ⁿC₃ a^n-3 b³
   ♦ We see that:
         ✰ The subscript of C is r
         ✰ The power of a is [n-r]
         ✰ The power of b is r
◼ So we get a new pattern in the subscript of C, power of a and power of b. Also, in this pattern, there is no (r-1). Only r.
5. Based on this, we can consider the (r+1)^th term as the general term.
• We get: (r+1)^th term = ⁿC_r a^n-r b^r
• The (r+1)^th term is denoted as: T_r+1
• So we can write: T_r+1 = ⁿC_r a^n-r b^r

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Some basics about middle term can be written in 8 steps:
1. We know that, if the index is n, then the number of terms will be (n+1)
• Also we know that, if n is even, (n+1) will be odd.
2. So we can write:
If the index n is even, there will be an odd number of terms in the expansion.
• For example, if the index is 6, there will be 7 terms in the expansion.
3. If the number of terms is odd, there will be a unique middle term. An example is shown in fig.8.4 below:

Fig.8.4

• There are a total of 7 terms. There are 3 terms on either sides. The fourth term is the middle term.
4. We can write a relation between the two items below:
   ♦ The index n
   ♦ The position of the middle term
• The relation is:
Position of the middle term when the index n is even = $\left(\frac{(n+1)+1}{2} \right)~=~\frac{n+2}{2}~=~\frac{n}{2}+1$
5. Next we will consider the case when n is odd.
• We know that if the index is n, then the number of terms will be (n+1)
• Also we know that if n is odd, (n+1) will be even.
6. So we can write:
If the index n is odd, there will be an even number of terms in the expansion.
• For example, if the index is 7, there will be 8 terms in the expansion.
7. If the number of terms is odd, there will be two middle terms. An example is shown in fig.8.5 below:

Fig.8.5

• There are a total of 8 terms. There are 3 terms on either sides. The fourth and fifth terms are the middle terms.
8. We can write a relation between the two items below:
   ♦ The index n
   ♦ The positions of the middle terms
• The relation is:
Positions of the middle terms when the index n is odd = $\frac{n+1}{2}~\text{and}~\frac{n+1}{2}+1$

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Now we know how to calculate the position of middle terms. Let us see an interesting case. It can be written in 5 steps:
1. We know that, if n is a natural number, 2n will be an even number.
2. So if the index is 2n, then there will be an odd number (2n+1) of terms.
3. We have seen that, if the number of terms is odd, there will be an unique middle term.
• Based on fig.8.4 above, the position of that middle term is: $\frac{2n}{2}+1~=~(n+1)$
4. Now consider the expansion of $\left(x+ \frac{1}{x} \right)^{2n}$
• We want to find the middle term, which is the (n+1)^th term.
• We know that, the (r+1)^th term of any binomial expansion is given by: ⁿC_r a^n-r b^r
5. So in our present case, the middle term, which is the (n+1)^th term will be:
${}^{2n} {\rm{C}}_n\;x^{2n-n}\;\left(\frac{1}{x} \right)^n$   
${}={}^{2n} {\rm{C}}_n\;x^{n}\;\left(\frac{1}{x} \right)^n$   
${}={}^{2n} {\rm{C}}_n$
• This term does not have x. So it is known as: the term independent of x.
• It is also known as the constant term.

Now we will see a solved example.

Solved example 8.5
Find a if the 17^th and 18^th terms of the expansion (2 + a)⁵⁰ are equal.
Solution:
1. We have the formula to find the (r+1)^th term: : T_r+1 = ⁿC_r a^n-r b^r
2. Put r = 16. We get:
17^th term = ${}^{50} {\rm{C}}_{16} × 2^{50-16} × a^{16}$   
= ${}^{50} {\rm{C}}_{16} × 2^{34} × a^{16}$   
3. Put r = 17. We get:
18^th term = ${}^{50} {\rm{C}}_18 × 2^{50-17} × a^17$   
= ${}^{50} {\rm{C}}_17 × 2^{33} × a^17$
4. Given that, the two terms are equal. So we can write:
${}^{50} {\rm{C}}_{16} × 2^{34} × a^{16}~=~{}^{50} {\rm{C}}_{17} × 2^{33} × a^{17}$
• This can be rearranged as: $\frac{{}^{50} {\rm{C}}_{16} × 2^{34}}{{}^{50} {\rm{C}}_{17} × 2^{33}}~=~\frac{a^{17}}{a^{16}}$       
$\Rightarrow \frac{{}^{50} {\rm{C}}_{16} × 2}{{}^{50} {\rm{C}}_{17}}~=~a$
• Thus we get: a = 1

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In the next section we will see a few more solved examples.

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