Chapter 9.7 - Miscellaneous Examples on chapter 9

In the previous section, we completed a discussion on sequences and series. In this section, we will see some miscellaneous examples.

Solved example 9.21
If p^th, q^th, r^th and s^th terms of an A.P are in G.P, then show that (p-q), (q-r), (r-s) are also in G.P.
Solution:
1. We want to prove that (p-q), (q-r), (r-s) are in G.P.
• That means, we have to prove that $\frac{q-r}{p-q}~=~\frac{r-s}{q-r}$

2. Let 'a' be the first term and 'd' the common difference of the A.P.
• Then the given terms can be written as:
    ♦ p^th term = a_p = a + (p-1)d
    ♦ q^th term = a_q = a + (q-1)d
    ♦ r^th term = a_r = a + (r-1)d
    ♦ s^th term = a_s = a + (s-1)d
3. Given that, the above four terms are in G.P. So we can write:

$\begin{array}{ll}
{}&{\frac{a_q}{a_p}}
&{~=~\frac{a_r}{a_q}}& {~=~\frac{a_s}{a_r}}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{\Rightarrow}&{\frac{a+(q-1)d}{a+(p-1)d}}
&{~=~\frac{a+(r-1)d}{a+(q-1)d}}& {~=~\frac{a+(s-1)d}{a+(r-1)d}}
&{\color {green} {\text{- - - - (a)}}}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{\Rightarrow}&{\frac{a+(q-1)d~-~[a+(p-1)d]}{a+(p-1)d}}
&{~=~\frac{a+(r-1)d~-~[a+(q-1)d]}{a+(q-1)d}}& {~=~\frac{a+(s-1)d~-~[a+(r-1)d]}{a+(r-1)d}}
&{\color {green} {\text{- - - - (b)}}}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{\Rightarrow}&{\frac{a+qd-d-a-pd+d}{a+(p-1)d}}
&{~=~\frac{a+rd-d-a-qd+d}{a+(q-1)d}}& {~=~\frac{a+sd-d-a-rd+d}{a+(r-1)d}}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{\Rightarrow}&{\frac{(q-p)d}{a+(p-1)d}}
&{~=~\frac{(r-q)d}{a+(q-1)d}}& {~=~\frac{(s-r)d}{a+(r-1)d}}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

\end{array}$

◼ Remarks:
(i) Line marked as (a):
This line is obtained by substituting the results from step (2)
(ii) Line marked as (b):
Here we apply componendo and dividendo rule:

$\begin{array}{ll}
{\text{If}}&{\frac{a}{b}~=~\frac{c}{d}}
&{\text{Then}}& {\frac{a-b}{b}~=~\frac{c-d}{d}}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{\text{Example:}}&{\frac{5}{2}~=~\frac{10}{4}}
&{\text{Then}}& {\frac{5-2}{2}~=~\frac{10-4}{4}~=~\frac{6}{4}~=~\frac{3}{2}}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

\end{array}$

4. Consider the above result in (3).
Take out the first two items. We get:

$\begin{array}{ll}
{}&{\frac{(q-p)d}{a+(p-1)d}}
&{~=~\frac{(r-q)d}{a+(q-1)d}}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{\Rightarrow}&{\frac{p-q}{a+(p-1)d}}
&{~=~\frac{q-r}{a+(q-1)d}}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{\Rightarrow}&{\frac{q-r}{p-q}}
&{~=~\frac{a+(q-1)d}{a+(p-1)d}}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

\end{array}$ 

5. Consider again the result in (3).
Take out the last two items. We get:

$\begin{array}{ll}
{}&{\frac{(r-q)d}{a+(q-1)d}}
&{~=~\frac{(s-r)d}{a+(r-1)d}}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{\Rightarrow}&{\frac{q-r}{a+(q-1)d}}
&{~=~\frac{r-s}{a+(r-1)d}}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{\Rightarrow}&{\frac{r-s}{q-r}}
&{~=~\frac{a+(r-1)d}{a+(q-1)d}}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

\end{array}$

• From the line marked as (a) in step (3), we get:
$\frac{a+(r-1)d}{a+(q-1)d}~=~\frac{a+(q-1)d}{a+(p-1)d}$

• So we can write: $\frac{r-s}{q-r}~=~\frac{a+(q-1)d}{a+(p-1)d}$

6. Comparing the results in (4) and (5), we get:

$\frac{q-r}{p-q}~=~\frac{a+(q-1)d}{a+(p-1)d}~=~\frac{r-s}{q-r}$

• Hence the statement in (1) is proved.

Solved example 9.22
If a, b, c are in G.P and $a^{\frac{1}{x}}~=~b^{\frac{1}{y}}~=~c^{\frac{1}{z}}$, prove that x, y, z are in A.P.
Solution:
1. We want to prove that x, y, z are in A.P.
• That means, we have to prove that y - x = z - y
2. Given that $a^{\frac{1}{x}}~=~b^{\frac{1}{y}}~=~c^{\frac{1}{z}}$
Let us assume that, the three quantities are equal to k. So we can write:
$a^{\frac{1}{x}}~=~b^{\frac{1}{y}}~=~c^{\frac{1}{z}}~=~k$
3. Consider the above result in (2). Taking the first and last items, we get:

$\begin{array}{ll}
{}&{a^{\frac{1}{x}}}
&{~=~k}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{\Rightarrow}&{\left(a^{\frac{1}{x}} \right)^x}
&{~=~k^x}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{\Rightarrow}&{\left(a^{\frac{x}{x}} \right)}
&{~=~k^x}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{\Rightarrow}&{\left(a^{1} \right)}
&{~=~k^x}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{\Rightarrow}&{a}
&{~=~k^x}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

\end{array}$

◼ In a similar way, we will get:
b = k^y and c = k^z

4. Now we get:
$\frac{b}{a}~=~\frac{k^y}{k^x}~=~k^{y-x}$
5. Also we get:
$\frac{c}{b}~=~\frac{k^z}{k^y}~=~k^{z-y}$
6. Given that a, b and c are in G.P. So we can write: $\frac{b}{a}~=~\frac{c}{b}$
7. So we can equate the results in (4) and (5). We get:
$\frac{b}{a}~=~\frac{c}{b}~=~k^{y-x}~=~k^{z-y}$
• Equating the powers of k, we get: y-x = z-y
• Hence the statement in (1) is proved.

Solved example 9.23
If a, b, c, d and p are different real numbers such that
$\left(a^2 + b^2 + c^2 \right)p^2~-~2(ab+bc+cd)p~+~\left(b^2 + c^2 + d^2 \right) ~\le~0$,
then show that a, b, c and d are in G.P.
Solution:
1. We want to prove that a, b, c and d are in G.P.
• That means, we have to prove that $\frac{b}{a}~=~\frac{c}{b}~=~\frac{d}{c}$.
2. The given inequality can be rearranged as follows:

$\begin{array}{ll}
{}&{\left(a^2 + b^2 + c^2 \right)p^2~-~2(ab+bc+cd)p~+~\left(b^2 + c^2 + d^2 \right)}
&{~\le~0}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{\Rightarrow}&{\left(a^2 p^2 - 2abp + b^2 \right)~+~\left(b^2 p^2 - 2bcp + c^2 \right)~+~\left(c^2 p^2 - 2cdp + d^2 \right)}
&{~\le~0}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{\Rightarrow}&{(ap-b)^2~+~(bp-c)^2~+~(cp-d)^2}
&{~\ge~0}& {\color {green} {\text{- - - - (a)}}}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

\end{array}$

◼ Remarks:
Line marked as (a):
• This line is a sum of three terms. Each of the three terms is a square.
    ♦ Given that: a, b, c, d and p are real numbers.
    ♦ So (ap-b), (bp-c) and (cp-d) will also be real numbers.
    ♦ Square of real numbers will be zero or +ve.
    ♦ So the sum of the three terms will be zero or +ve.
    ♦ Thus the sign in this line is changed to ≥.

3. The given inequality is only rearranged in the above step (2).
• The rearranged result in (2) is same as the given inequality.
• But the given inequality is ≤. In step (2), it changed to ≥.
• If both ≤ and ≥ are to be satisfied, the only possibility is that, both the expressions are equal to zero. 
4. So we can write: $(ap-b)^2~+~(bp-c)^2~+~(cp-d)^2~=~0$
• If the sum of certain squares is zero, each term in that sum must be zero.
• So we get:
(i) (ap-b)² = 0
(ii) (bp-c)² = 0
(iii) (cp-d)² = 0
5. If square of a real number is zero, then that real number must be zero.
So we get:
(i) ap-b = 0
(ii) bp-c = 0
(iii) cp-d = 0
• From 5(i), we get: $p=\frac{b}{a}$
• From 5(ii), we get: $p=\frac{c}{b}$
• From 5(iiii), we get: $p=\frac{d}{c}$
6. Based on the above step (5), we can write:
$\frac{b}{a}~=~\frac{c}{b}~=~\frac{d}{c}~=~p$
• Hence the statement in (1) is proved.

Solved example 9.24
If p, q, r are in G.P and the equations px² + 2qx + r = 0 and dx² + 2ex + f = 0 have a common root, then show that $\frac{d}{p}, \frac{e}{q}, \frac{f}{r}$ are in A.P.
Solution:
1. We want to prove that $\frac{d}{p}, \frac{e}{q}, \frac{f}{r}$ are in A.P.
• That means, we have to prove that:

$\begin{array}{ll}
{}&{\frac{e}{q} - \frac{d}{p}}
&{~=~\frac{f}{r} - \frac{e}{q}}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{\Rightarrow}&{\frac{e}{q} + \frac{e}{q}}
&{~=~\frac{f}{r} + \frac{d}{p}}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{\Rightarrow}&{\frac{2e}{q}}
&{~=~\frac{f}{r} + \frac{d}{p}}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

\end{array}$

2. Given that, p, q, r are in G.P. So we can write: q² = pr
3. Also given that, px² + 2qx + r = 0 and dx² + 2ex + f = 0 have a common root.
• Let us write the roots of the first equation. We get:

$\begin{array}{ll}
{}&{\frac{-2q \pm \sqrt{(2q)^2 ~-~4pr}}{2p}}
&{~=~\frac{-2q \pm \sqrt{4q^2 ~-~4pr}}{2p}}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{~=~}&{\frac{-2q \pm \sqrt{4pr ~-~4pr}}{2p}}
&{~=~\frac{-2q \pm \sqrt{0}}{2p}}& {\color {green} {\text{- - - - (a)}}}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{~=~}&{\frac{-2q}{2p}}
&{~=~\frac{-q}{p}}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

\end{array}$

◼ Remarks:
Line marked as (a): From (2), we have: q² = pr

4. So the first equation has only one root, which is: $\frac{-q}{p}$
• Given that, the two equations have a common root. So $\frac{-q}{p}$must be a root of the second equation also.
• Since $\frac{-q}{p}$ is a root of the second equation, we can write:

$\begin{array}{ll}
{}&{d \left(\frac{-q}{p} \right)^2~+~2e\left(\frac{-q}{p} \right)~+~f}
&{~=~0}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{\Rightarrow}&{\frac{d q^2}{p^2}~-~\frac{2eq}{p}~+~f}
&{~=~0}& {}
&{\color {green} {\text{- - - - (a)}}}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{\Rightarrow}&{\frac{d q^2}{p^2 q^2}~-~\frac{2eq}{p q^2}~+~\frac{f}{q^2}}
&{~=~\frac{d}{p^2}~-~\frac{2e}{p q}~+~\frac{f}{q^2}}& {~=~0}
&{\color {green} {\text{- - - - (b)}}}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{\Rightarrow}&{\frac{p d}{p^2}~-~\frac{2ep}{p q}~+~\frac{f p}{q^2}}
&{~=~\frac{d}{p}~-~\frac{2e}{q}~+~\frac{f p}{q^2}}& {~=~0}
&{\color {green} {\text{- - - - (c)}}}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{\Rightarrow}&{\frac{d}{p}~-~\frac{2e}{q}~+~\frac{f p}{pr}}
&{~=~\frac{d}{p}~-~\frac{2e}{q}~+~\frac{f}{r}}& {~=~0}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{\Rightarrow}&{\frac{2e}{q}}
&{~=~\frac{d}{p}~+~\frac{f}{r}}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

\end{array}$

◼ Remarks:
(i) Line marked as (a):
We want to transform the statement in (a) to the staement in (1). For that, we divide by q²
(ii) Line marked as (b):
We want to further transform the statement in (b) to the statement in (1). For that, we multiply by p.
(iii) Line marked as (c):
From (2), we have: q² = pr
• Hence the statement in (1) is proved.

---

The link below gives some more solved examples

Miscellaneous Exercise on chapter 9

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In the next chapter we will see straight lines.

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Chapter 9.6 - Sum of The Cubes of First n Natural Numbers

In the previous section, we saw the sum of squares of the first n natural numbers. In this section, we will see sum of cubes of the first n natural numbers. 

C. 1³ + 2³ + 3³ +. . . + n³

This sum can be calculated in 8 steps:
1. Consider the identity: (a+b)⁴ = a⁴+4a³b+6a²b²+4ab³+b⁴
[Recall that (a+b) can be raised to any power by using binomial theorem that we saw in the previous chapter]
Let us put a = k and b = -1. We get:

$\begin{array}{ll}
{}&{[k+(-1)]^4}
&{}={}& {[k-1]^4}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{}&{}
&{}={}& {k^4 ~+~ 4 × k^3 × -1 ~+~6 × k^2 × (-1)^2~+~4 × k × (-1)^3~+~(-1)^4}
&{}& {}
&{}{}& {}
&{}& {}&{}& {} &{} \\


{}&{}
&{}={}& {k^4 ~-~ 4 k^3~+~6k^2 ~-~4k~+~1}
&{}& {}
&{}{}& {}
&{}& {}&{}& {} &{} \\

\end{array}$

2. Subtracting (k-1)⁴ from k⁴, we get:

$\begin{array}{ll}
{}&{k^4~-~(k-1)^4}
&{}={}& {k^4~-~\left(k^4 ~-~ 4 k^3~+~6k^2 ~-~4k~+~1 \right)}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{}&{}
&{}={}& {k^4~-~k^4~+~4 k^3~-~6k^2 ~+~4k~-~1}
&{}& {}
&{}{}& {}
&{}& {}&{}& {} &{} \\

{}&{}
&{}={}& {4 k^3~-~6k^2 ~+~4k~-~1}
&{}& {}
&{}{}& {}
&{}& {}&{}& {} &{} \\
\end{array}$

• We can use this as an identity:
$k^4~-~(k-1)^4~=~4 k^3~-~6k^2 ~+~4k~-~1$

3. In the above identity, let us put k = 1, 2, 3, . . . , n successively. We get:

$\begin{array}{ll}
{\text{When k = 1,}}&{1^4~-~(1-1)^4}
&{}={}& {1^4~-~0^4}
&{}={}& {4 × 1^3~-~6 × 1^2 ~+~4 × 1~-~1}
&{}={}& {4(1)^3~-~6(1)^2~+~4(1)~-~1}
&{}& {}&{}& {} &{} \\

{\text{When k = 2,}}&{2^4~-~(2-1)^4}
&{}={}& {2^4~-~1^4}
&{}={}& {4 × 2^3~-~6 × 2^2 ~+~4 × 2~-~1}
&{}={}& {4(2)^3~-~6(2)^2~+~4(2)~-~1}
&{}& {}&{}& {} &{} \\

{\text{When k = 3,}}&{3^4~-~(3-1)^4}
&{}={}& {3^4~-~2^4}
&{}={}& {4 × 3^3~-~6 × 3^2 ~+~4 × 3~-~1}
&{}={}& {4(3)^3~-~6(3)^2~+~4(3)~-~1}
&{}& {}&{}& {} &{} \\

{\text{When k = 4,}}&{4^4~-~(4-1)^4}
&{}={}& {4^4~-~3^4}
&{}={}& {4 × 4^3~-~6 × 4^2 ~+~4 × 4~-~1}
&{}={}& {4(4)^3~-~6(4)^2~+~4(4)~-~1}
&{}& {}&{}& {} &{} \\


{-}&{-}
&{}& {-}
&{}& {-}
&{}& {-}
&{}& {}&{}& {} &{} \\

{-}&{-}
&{}& {-}
&{}& {-}
&{}& {-}
&{}& {}&{}& {} &{} \\

{\text{When k = n,}}&{n^4~-~(n-1)^4}
&{}& {}
&{}={}& {4 × n^3~-~6 × n^2 ~+~4 × n~-~1}
&{}={}& {4(n)^3~-~6(n)^2~+~4(n)~-~1}
&{}& {}&{}& {} &{} \\

\end{array}$

4. Picking the first and last items from each line, we get:

$\begin{array}{ll}
{}&{}
&{}& {1^4~-~0^4}
&{}& {}
&{}={}& {4(1)^3~-~6(1)^2~+~4(1)~-~1}
&{}& {}&{}& {} &{} \\

{}&{}
&{}& {2^4~-~1^4}
&{}& {}
&{}={}& {4(2)^3~-~6(2)^2~+~4(2)~-~1}
&{}& {}&{}& {} &{} \\

{}&{}
&{}& {3^4~-~2^4}
&{}& {}
&{}={}& {4(3)^3~-~6(3)^2~+~4(3)~-~1}
&{}& {}&{}& {} &{} \\

{}&{}
&{}& {4^4~-~3^4}
&{}& {}
&{}={}& {4(4)^3~-~6(4)^2~+~4(4)~-~1}
&{}& {}&{}& {} &{} \\

{}&{}
&{}& {-}
&{}& {}
&{}& {-}
&{}& {}&{}& {} &{} \\

{}&{}
&{}& {-}
&{}& {}
&{}& {-}
&{}& {}&{}& {} &{} \\

{}&{}
&{}& {n^4~-~(n-1)^4}
&{}& {}
&{}={}& {4(n)^3~-~6(n)^2~+~4(n)~-~1}
&{}& {}&{}& {} &{} \\

\end{array}$

5. In the above result in (4), let us add all terms on the left side.
• We see that diagonal elements will get cancelled:
   ♦ 1⁴ will get cancelled by -1⁴.
   ♦ 2⁴ will get cancelled by -2⁴.
   ♦ 3⁴ will get cancelled by -3⁴.
   ♦ so on . . .
• So only 0⁴ and n⁴ will remain.
• Thus the sum of all terms on the left side is: (n⁴ - 0⁴) = n⁴.

6. In the result in (4), let us add all terms on the right side. We get:
4(1³ + 2³ + 3³ +. . . + n³) - 6(1² + 2² + 3² +. . . + n²) + 4(1 + 2 + 3 +. . . + n) - (1+1+1+ . . . n times)
• This can be written in a shortened form using sigma notations:
$4 \sum\limits_{k\,=\,0}^{k\,=\,n}{k^3}~-~6 \sum\limits_{k\,=\,0}^{k\,=\,n}{k^2}~+~4 \sum\limits_{k\,=\,0}^{k\,=\,n}{k}~-~n$

7. Equating the results in (5) and (6), we get:

$\begin{array}{ll}
{}&{n^4}
&{}={}& {4 \sum\limits_{k\,=\,0}^{k\,=\,n}{k^3}~-~6 \sum\limits_{k\,=\,0}^{k\,=\,n}{k^2}~+~4 \sum\limits_{k\,=\,0}^{k\,=\,n}{k}~-~n}
&{}& {}
 \\

{\Rightarrow}&{4 \sum\limits_{k\,=\,0}^{k\,=\,n}{k^3}}
&{}={}& {n^4~+~6 \sum\limits_{k\,=\,0}^{k\,=\,n}{k^2}~-~4 \sum\limits_{k\,=\,0}^{k\,=\,n}{k}~+~n}
&{\color {green} {\text{- - - - (a)}}}& {}
 \\

{\Rightarrow}&{4 \sum\limits_{k\,=\,0}^{k\,=\,n}{k^3}}
&{}={}& {n^4~+~\left[\frac{6n(2n + 1) (n+1)}{6} \right]~-~4 \sum\limits_{k\,=\,0}^{k\,=\,n}{k}~+~n}
&{\color {green} {\text{- - - - (b)}}}& {}
 \\

{\Rightarrow}&{4 \sum\limits_{k\,=\,0}^{k\,=\,n}{k^3}}
&{}={}& {n^4~+~\left[\frac{6n(2n + 1) (n+1)}{6} \right]~-~\left[\frac{4n(n+1)}{2} \right]~+~n}
&{}& {}
 \\

{\Rightarrow}&{4 \sum\limits_{k\,=\,0}^{k\,=\,n}{k^3}}
&{}={}& {n^4~+~n(2n + 1) (n+1)~-~2n(n+1)~+~n}
&{}& {}
 \\

{\Rightarrow}&{4 \sum\limits_{k\,=\,0}^{k\,=\,n}{k^3}}
&{}={}& {n^4~+~2n^3+2n^2+n^2+n~-~2n^2 -2n~+~n}
&{}& {}
 \\

{\Rightarrow}&{4 \sum\limits_{k\,=\,0}^{k\,=\,n}{k^3}}
&{}={}& {n^4 + 2n^3 + n^2 }
&{}& {}
 \\

{\Rightarrow}&{4 \sum\limits_{k\,=\,0}^{k\,=\,n}{k^3}}
&{}={}& {n^2 \left(n^2 + 2n + 1 \right) }
&{\color {green} {\text{- - - - (c)}}}& {}
 \\

{\Rightarrow}&{4 \sum\limits_{k\,=\,0}^{k\,=\,n}{k^3}}
&{}={}& {n^2 \left(n + 1 \right)^2 }
&{}& {}
 \\

{\Rightarrow}&{\sum\limits_{k\,=\,0}^{k\,=\,n}{k^3}}
&{}={}& {\frac{n^2 \left(n + 1 \right)^2}{4} }
&{}& {}
 \\

{\Rightarrow}&{\sum\limits_{k\,=\,0}^{k\,=\,n}{k^3}}
&{}={}& {\left[ \frac{n (n + 1)}{2}\right]^2}
&{}& {}
 \\
\end{array}$

Remarks:
(i) The line marked as (a):
• $\sum\limits_{k\,=\,0}^{k\,=\,n}{k^2}$ is in fact the sum of squares of first n natural numbers.
• So our second result B that we saw in the previous section can be used.
(ii) The line marked as (b):
• $\sum\limits_{k\,=\,0}^{k\,=\,n}{k}$ is in fact the sum of first n natural numbers.
• So our first result A that we saw in the previous section can be used.
(iii) The line marked as (c):
• Using the identity (a+b)² = a² + 2ab + b²,
(n² + 2n + 1) is (n+1)²

8. We can write the result as a formula:
Sum of cubes of the first n natural numbers
= 1³ + 2³ + 3³ +. . . + n³

= $\sum\limits_{k\,=\,1}^{k\,=\,n}{k^3}~=~\left[ \frac{n (n + 1)}{2}\right]^2$

---

Let us see some solved examples:

Solved example 9.19
Find the sum to n terms of the series: 5 + 11 + 19 + 29 + 41 + . . . 
Solution:
1. Let us write:
$S_n~=~5~+~11~+~19~+~29~+~41~+~.~.~.~+~a_{n-1}~+~a_{n}$
2. The same result can be written as:
$S_n~=~5~+~11~+~19~+~29~+~41~+~.~.~.~+~a_{n-2}~+~a_{n-1}~+~a_{n}$
3. We will write the second result just below the first result. But one term to the right. We get:

$\begin{array}{ll}
{S_n}&{}={}
&{5}& {~+~}
&{11}& {~+~}
&{19}& {~+~}
&{29}& {~+~.~.~.~+~}&{a_{n-1}}& {~+~} &{a_n} &{} &{} \\

{S_n}&{}={}
&{}& {}
&{5}& {~+~}
&{11}& {~+~}
&{19}& {~+~.~.~.~+~}&{a_{n-2}}& {~+~} &{a_{n-1}} &{~+~} &{a_n} \\

\end{array}$

4. Subtracting each term in the second row, from the term directly above it, we get:

$\begin{array}{ll}
{S_n}&{}={}
&{5}& {~+~}
&{11}& {~+~}
&{19}& {~+~}
&{29}& {~+~.~.~.~+~}&{a_{n-1}}& {~+~} &{a_n} &{} &{} \\

{S_n}&{}={}
&{}& {}
&{5}& {~+~}
&{11}& {~+~}
&{19}& {~+~.~.~.~+~}&{a_{n-2}}& {~+~} &{a_{n-1}} &{~+~} &{a_n} \\

{0}&{}={}
&{5}& {~+~}
&{\left\{6 \right. }& {~+~}
&{8}& {~+~}
&{10}& {~+~.~.~.~}&{}& {} &{\left. \right \}} &{~-~} &{a_n} \\

\end{array}$

5. In the above result, there will be (n-1) terms inside the curly brackets '{}'
• In side the curly brackets, we have the series: 6 + 8 + 10 + . . . (n-1) terms
• It is an A.P with a = 6, d = 2 and n = (n-1)
• So sum of the series = $\frac{(n-1)}{2} [2 × 6 + (n-1-1)2]~=~\frac{(n-1)}{2} [12 + (n-2)2]~=~(n-1)(6+n-2)~=~(n-1)(4+n)$

6. Now the result in (4) becomes:
0 = 5 + (n-1)(4+n) - an
⇒ a_n = 5 + (n-1)(4+n)
⇒ a_n = 5 + 4n + n² - 4 - n
⇒ a_n = 1 + 3n + n²
⇒ a_n = n² + 3n + 1

7. Thus we obtained the n^th term.
• n^th term = a_n = n² + 3n + 1.
• So the sum (S_n) of the given series can be obtained as:

$\begin{array}{ll}
{S_n}&{}={}
&{\sum\limits_{k\,=\,0}^{k\,=\,n}{a_k}}~=~\sum\limits_{k\,=\,0}^{k\,=\,n}{\left(k^2 + 3k + 1 \right)}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{}&{}={}
&{\sum\limits_{k\,=\,0}^{k\,=\,n}{k^2}~+~3\sum\limits_{k\,=\,0}^{k\,=\,n}{k}~+~n}& {}
&{\color {green} {\text{- - - - (a)}}}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{}&{}={}
&{\frac{n(2n + 1) (n+1)}{6}~+~3\sum\limits_{k\,=\,0}^{k\,=\,n}{k}~+~n}& {}
&{\color {green} {\text{- - - - (b)}}}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{}&{}={}
&{\frac{n(2n + 1) (n+1)}{6}~+~ \frac{3n (n+1)}{2}~+~n}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{}&{}={}
&{\frac{n(2n + 1) (n+1)~+~9n(n+1)~+~6n}{6}}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{}&{}={}
&{\frac{n[(2n+1)(n+1)+9(n+1)+6]}{6}}& {}={}
&{\frac{n[2n^2+2n+n+1+9n+9+6]}{6}}& {}={}
&{\frac{n[2n^2+12n+16]}{6}}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{}&{}={}
&{\frac{2n[n^2+6n+8]}{6}}& {}={}
&{\frac{n[n^2+6n+8]}{3}}& {}
&{\color {green} {\text{- - - - (c)}}}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{}&{}={}
&{\frac{n[(n+2)(n+4)]}{3}}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

\end{array}$

Remarks:
(i) The line marked as (a):
• $\sum\limits_{k\,=\,0}^{k\,=\,n}{k^2}$ is in fact the sum of squares of first n natural numbers.
• So our second result B that we saw in the previous section can be used.
(ii) The line marked as (b):
• $\sum\limits_{k\,=\,0}^{k\,=\,n}{k}$ is in fact the sum of first n natural numbers.
• So our first result A that we saw in the previous section can be used.
(iii) The line marked as (c):
• n² + 6n + 8 = 0 can be solved as a quadratic equation.
• We will get: n = -2 and n = -4.

Solved example 9.20
Find the sum to n terms of the series whose n^th term is n(n+3)
Solution:
• n^th term = n(n+3).
• So the sum (S_n) of the given series can be obtained as:

$\begin{array}{ll}
{S_n}&{}={}
&{\sum\limits_{k\,=\,0}^{k\,=\,n}{[k(k+3)]}}~=~\sum\limits_{k\,=\,0}^{k\,=\,n}{\left[k^2 + 3k \right]}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{}&{}={}
&{\sum\limits_{k\,=\,0}^{k\,=\,n}{k^2}~+~3\sum\limits_{k\,=\,0}^{k\,=\,n}{k}}& {}
&{\color {green} {\text{- - - - (a)}}}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{}&{}={}
&{\frac{n(2n + 1) (n+1)}{6}~+~3\sum\limits_{k\,=\,0}^{k\,=\,n}{k}}& {}
&{\color {green} {\text{- - - - (b)}}}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{}&{}={}
&{\frac{n(2n + 1) (n+1)}{6}~+~ \frac{3n (n+1)}{2}}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{}&{}={}
&{\frac{n(2n + 1) (n+1)~+~9n(n+1)}{6}}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{}&{}={}
&{\frac{n[(2n+1)(n+1)+9(n+1)]}{6}}& {}={}
&{\frac{n[2n^2+2n+n+1+9n+9]}{6}}& {}={}
&{\frac{n[2n^2+12n+10]}{6}}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{}&{}={}
&{\frac{2n[n^2+6n+5]}{6}}& {}={}
&{\frac{n[n^2+6n+5]}{3}}& {}
&{\color {green} {\text{- - - - (c)}}}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{}&{}={}
&{\frac{n[(n+1)(n+5)]}{3}}& {}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

\end{array}$

Remarks:
(i) The line marked as (a):
• $\sum\limits_{k\,=\,0}^{k\,=\,n}{k^2}$ is in fact the sum of squares of first n natural numbers.
• So our second result B that we saw in the previous section can be used.
(ii) The line marked as (b):
• $\sum\limits_{k\,=\,0}^{k\,=\,n}{k}$ is in fact the sum of first n natural numbers.
• So our first result A that we saw in the previous section can be used.
(iii) The line marked as (c):
• n² + 6n + 5 = 0 can be solved as a quadratic equation.
• We will get: n = -1 and n = -5.

---

The link below gives some more solved examples

Exercise 9.4

---

In the next section we will see some miscellaneous examples.

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Chapter 9.5 - Sum of The Squares of First n Natural Numbers

In the previous section, we completed a discussion on geometric progression and geometric mean. In this section, we will see sum to n terms of special series.

We have to find the sum of three series. They are:

A. 1 + 2 + 3 +. . . + n
   ♦ This is the sum of first n natural numbers.

B. 1² + 2² + 3² +. . . + n²
   ♦ This is the sum of squares of the first n natural numbers.

C. 1³ + 2³ + 3³ +. . . + n³
   ♦ This is the sum of cubes of the first n natural numbers.

---

A. 1 + 2 + 3 +. . . + n

This sum can be calculated in 3 steps:
1. The given series is the series related to the sequence 1, 2, 3, . . . , n
2. This sequence is an A.P with a = 1 and d = 1
• So sum to n terms will be given by:
$\begin{array}{ll}
{}&{S_n}
&{}={}& {\frac{n}{2}[2a+(n-1)d]}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} \\

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


{}&{}
&{}={}& {\frac{n}{2}[2+n-1]}
&{}& {}
&{}{}& {}
&{}& {}&{}& {} &{} \\


{}&{}
&{}={}& {\frac{n}{2}[n+1]}
&{}& {}
&{}{}& {}
&{}& {}&{}& {} &{} \\
\end{array}$

3. We can write it as a formula:
Sum of first n natural numbers = $\frac{n}{2}[n+1]$

---

B. 1² + 2² + 3² +. . . + n²

This sum can be calculated in 8 steps:
1. Consider the identity: (a+b)³ = a³ + 3a²b + 3 ab² + b³
Let us put a = k and b = -1. We get:

$\begin{array}{ll}
{}&{[k+(-1)]^3}
&{}={}& {[k-1]^3}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{}&{}
&{}={}& {k^3 ~+~ 3 × k^2 × -1 ~+~3 × k × (-1)^2~+~(-1)^3}
&{}& {}
&{}{}& {}
&{}& {}&{}& {} &{} \\


{}&{}
&{}={}& {k^3 ~-~ 3 k^2 ~+~3k~-~1}
&{}& {}
&{}{}& {}
&{}& {}&{}& {} &{} \\

\end{array}$

2. Subtracting (k-1)³ from k³, we get:

$\begin{array}{ll}
{}&{k^3~-~(k-1)^3}
&{}={}& {k^3~-~\left(k^3 ~-~ 3 k^2 ~+~3k~-~1 \right)}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{}&{}
&{}={}& {k^3~-~k^3 ~+~ 3 k^2 ~-~3k~+~1}
&{}& {}
&{}{}& {}
&{}& {}&{}& {} &{} \\

{}&{}
&{}={}& {3 k^2 ~-~3k~+~1}
&{}& {}
&{}{}& {}
&{}& {}&{}& {} &{} \\
\end{array}$

• We can use this as an identity:
$k^3~-~(k-1)^3~=~3 k^2 ~-~3k~+~1$

3. In the above identity, let us put k = 1, 2, 3, . . . , n successively. We get:

$\begin{array}{ll}
{\text{When k = 1,}}&{1^3~-~(1-1)^3}
&{}={}& {1^3~-~0^3}
&{}={}& {3 × 1^2 ~-~3 × 1~+~1}
&{}={}& {3(1)^2~-~3(1)~+~1}
&{}& {}&{}& {} &{} \\

{\text{When k = 2,}}&{2^3~-~(2-1)^3}
&{}={}& {2^3~-~1^3}
&{}={}& {3 × 2^2 ~-~3 × 2~+~1}
&{}={}& {3(2)^2~-~3(2)~+~1}
&{}& {}&{}& {} &{} \\

{\text{When k = 3,}}&{3^3~-~(3-1)^3}
&{}={}& {3^3~-~2^3}
&{}={}& {3 × 3^2 ~-~3 × 3~+~1}
&{}={}& {3(3)^2~-~3(3)~+~1}
&{}& {}&{}& {} &{} \\

{\text{When k = 4,}}&{4^3~-~(4-1)^3}
&{}={}& {4^3~-~3^3}
&{}={}& {3 × 4^2 ~-~3 × 4~+~1}
&{}={}& {3(4)^2~-~3(4)~+~1}
&{}& {}&{}& {} &{} \\

{-}&{-}
&{}& {-}
&{}& {-}
&{}& {-}
&{}& {}&{}& {} &{} \\

{-}&{-}
&{}& {-}
&{}& {-}
&{}& {-}
&{}& {}&{}& {} &{} \\

{\text{When k = n,}}&{n^3~-~(n-1)^3}
&{}& {}
&{}={}& {3 × n^2 ~-~3 × n~+~1}
&{}={}& {3(n)^2~-~3(n)~+~1}
&{}& {}&{}& {} &{} \\

\end{array}$

4. Picking the first and last items from each line, we get:

$\begin{array}{ll}
{}&{}
&{}& {1^3~-~0^3}
&{}& {}
&{}={}& {3(1)^2~-~3(1)~+~1}
&{}& {}&{}& {} &{} \\

{}&{}
&{}& {2^3~-~1^3}
&{}& {}
&{}={}& {3(2)^2~-~3(2)~+~1}
&{}& {}&{}& {} &{} \\

{}&{}
&{}& {3^3~-~2^3}
&{}& {}
&{}={}& {3(3)^2~-~3(3)~+~1}
&{}& {}&{}& {} &{} \\

{}&{}
&{}& {4^3~-~3^3}
&{}& {}
&{}={}& {3(4)^2~-~3(4)~+~1}
&{}& {}&{}& {} &{} \\

{}&{}
&{}& {-}
&{}& {}
&{}& {-}
&{}& {}&{}& {} &{} \\

{}&{}
&{}& {-}
&{}& {}
&{}& {-}
&{}& {}&{}& {} &{} \\

{}&{}
&{}& {n^3~-~(n-1)^3}
&{}& {}
&{}={}& {3(n)^2~-~3(n)~+~1}
&{}& {}&{}& {} &{} \\

\end{array}$

5. In the above result in (4), let us add all terms on the left side.
• We see that diagonal elements will get cancelled:
   ♦ 1³ will get cancelled by -1³.
   ♦ 2³ will get cancelled by -2³.
   ♦ 3³ will get cancelled by -3³.
   ♦ so on . . .
• So only 0³ and n³ will remain.
• Thus the sum of all terms on the left side is: (n³ - 0³) = n³.

6. In the result in (4), let us add all terms on the right side. We get:
3(1² + 2² + 3² +. . . + n²) - 3(1 + 2 + 3 +. . . + n) + (1+1+1+ . . . n times)
• This can be written in a shortened form using sigma notations:
$3 \sum\limits_{k\,=\,0}^{k\,=\,n}{k^2}~-~3 \sum\limits_{k\,=\,0}^{k\,=\,n}{k}~+~n$

7. Equating the results in (5) and (6), we get:

$\begin{array}{ll}
{}&{n^3}
&{}={}& {3 \sum\limits_{k\,=\,1}^{k\,=\,n}{k^2}~-~3 \sum\limits_{k\,=\,0}^{k\,=\,n}{k}~+~n}
&{}& {}
 \\

{\Rightarrow}&{3 \sum\limits_{k\,=\,1}^{k\,=\,n}{k^2}}
&{}={}& {n^3~+~3 \sum\limits_{k\,=\,0}^{k\,=\,n}{k}~-~n}
&{\color {green} {\text{- - - - (a)}}}& {}
 \\

{\Rightarrow}&{3 \sum\limits_{k\,=\,1}^{k\,=\,n}{k^2}}
&{}={}& {n^3~+~\frac{3n(n+1)}{2}~-~n}
&{}& {}
 \\

{\Rightarrow}&{3 \sum\limits_{k\,=\,1}^{k\,=\,n}{k^2}}
&{}={}& {\frac{2n^3 +3n(n+1) - 2n}{2}}
&{}& {}
 \\

{\Rightarrow}&{3 \sum\limits_{k\,=\,1}^{k\,=\,n}{k^2}}
&{}={}& {\frac{2n^3 + 3n^2 + 3n - 2n}{2}}
&{}& {}
 \\

{}&{}
&{}& {}
&{}& {}
 \\

{\Rightarrow}&{\sum\limits_{k\,=\,1}^{k\,=\,n}{k^2}}
&{}={}& {\frac{2n^3 + 3n^2 + n}{6}}
&{}& {}
 \\

{}&{}
&{}& {}
&{}& {}
 \\

{\Rightarrow}&{\sum\limits_{k\,=\,1}^{k\,=\,n}{k^2}}
&{}={}& {\frac{2n^3 + 3n^2 + n}{6}}
&{}& {}
 \\

{}&{}
&{}& {}
&{}& {}
 \\

{\Rightarrow}&{\sum\limits_{k\,=\,1}^{k\,=\,n}{k^2}}
&{}={}& {\frac{n(2n^2 + 3n + 1)}{6}}
&{\color {green} {\text{- - - - (b)}}}& {}
 \\

{}&{}
&{}& {}
&{}& {}
 \\

{\Rightarrow}&{\sum\limits_{k\,=\,1}^{k\,=\,n}{k^2}}
&{}={}& {\frac{n(2n + 1) (n+1)}{6}}
&{}& {}
\\

\end{array}$

Remarks:
(i) The line marked as (a):
• $\sum\limits_{k\,=\,0}^{k\,=\,n}{k}$ is in fact the sum of first n natural numbers.
• So our first result A that we saw at the beginning of this section can be used.
(ii) The line marked as (b):
• (2n² + 3n +1) can be written as (2n² + 2n + n + 1)
• But (2n² + 2n + n + 1) = [2n(n+1) + (n+1)] = [(2n+1)(n+1)]
So we get: (2n² + 3n + 1) = (2n+1)(n+1)

Another method:
• Find the solutions of the quadratic equation 2n² + 3n + 1 = 0
• The solutions are: n = $-\frac{1}{2}$ and n = -1
• So we can write: $2n^2 + 3n + 1 ~=~ \left(n+ \frac{1}{2} \right) (n+1) ~=~0$
$\Rightarrow~\left(\frac{2n+1}{2} \right) (n+1) ~=~0$
$\Rightarrow~\left(2n+1 \right) (n+1) ~=~0$

8. We can write it as a formula:
Sum of squares of the first n natural numbers
= 1² + 2² + 3² +. . . + n²

= $\sum\limits_{k\,=\,1}^{k\,=\,n}{k^2}~=~\frac{n(2n + 1) (n+1)}{6}$

---

In the next section we will see sum of the cubes of first n natural numbers.

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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}$

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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.

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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) $\sum\limits_{k\,=\,1}^{k\,=\,\infty}{2k-1}$

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

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