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 + . . .

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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:
$\begin{array}{ll} {\text{When n = 1,}~\frac{a_{n+1}}{a_n}}&{}={} &{\frac{a_{1+1}}{a_1}}& {}={} &{\frac{a_{2}}{a_1}}& {}={} &{\frac{1}{1}}& {} &{}& {}&{}& {} &{} \\ {\text{When n = 2,}~\frac{a_{n+1}}{a_n}}&{}={} &{\frac{a_{2+1}}{a_2}}& {}={} &{\frac{a_{3}}{a_2}}& {}={} &{\frac{2}{1}}& {} &{}& {}&{}& {} &{} \\ {\text{When n = 3,}~\frac{a_{n+1}}{a_n}}&{}={} &{\frac{a_{3+1}}{a_3}}& {}={} &{\frac{a_{4}}{a_3}}& {}={} &{\frac{3}{2}}& {} &{}& {}&{}& {} &{} \\ {\text{When n = 4,}~\frac{a_{n+1}}{a_n}}&{}={} &{\frac{a_{4+1}}{a_4}}& {}={} &{\frac{a_{5}}{a_4}}& {}={} &{\frac{5}{3}}& {} &{}& {}&{}& {} &{} \\ {\text{When n = 5,}~\frac{a_{n+1}}{a_n}}&{}={} &{\frac{a_{5+1}}{a_5}}& {}={} &{\frac{a_{6}}{a_5}}& {}={} &{\frac{8}{5}}& {} &{}& {}&{}& {} &{} \\ \end{array}$

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In the next section we will see arithmetic progression.

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