Sum of series

In this section, we intend to discuss sum of series or the sum of 'n' terms of special series. Example: series of natural numbers, series of square of natural numbers, series of cubes of natural numbers etc.
1) Sum of first 'n' natural numbers :
1 + 2 + 3 + ... + n = $\frac{n(n + 1)}{2}$

2) Sum of the squares of first 'n' natural numbers : $1^{2} + 2^{2} + 3^{2} + ...+ n^{2} = \frac{n(n + 1)(2n + 1)}{6}$

3) Sum of the cubes of first 'n' natural numbers : $1^{3} + 2^{3} + 3^{3} + ...+ n^{3} = \left \{ \frac{n(n+1)}{2} \right \}^{2}$

4) In series $a_{1} + a_{2} + a_{3} + ...+ a_{n}$
(i) if the differences $a_{2} - a_{1}, a_{3} - a_{2}$ ,... are in A.P then the nth term is given by
$a_{n} = an^{2}$ + bn + c, where a, b,c are constants.

(ii) if the differences $a_{2} - a_{1}, a_{3} - a_{2}$ ,... are in G.P then the nth term is given by
$a_{n} = ar^{n - 1} + b_{n}$ + c, where a, b,c are constants.

Summation of some special series :
$\frac{1}{a(a+d))} + \frac{1}{(a+d)(a + 2d)} + ...+ \frac{1}{(a+(n-2)d)(a + (n-1)d)} = \frac{n - 1}{a +(n - 1)d}$

Examples on sum of series

Find the sum of of the following series :
(i) $2^{2} + 4^{2} + 6^{2} + 8^{2}$ + ...
The nth term will be
$a_{n}$ = a + (n - 1)d
a = 2, d = 2
$a_{n} = [2 + (n - 1)2]^{2}$
$a_{n} = (2n)^{2}$
$a_{n} = 4n^{2}$
$2^{2} + 4^{2} + 6^{2} + 8^{2}$ + ... to n terms
$\sum_{k = 2}^{n}{T_{k}} = \sum_{k = 2}^{k} 4k^{2}$

= $\sum_{k = 2}^{n} 4k^{2} $

= 4 $\sum_{k = 2}^{n} k^{2} $

= 4$\frac{n(n + 1)(2n + 1)}{6}$
= 2$\frac{n(n + 1)(2n + 1)}{3}$
= 4 $\frac{n(n + 1)(2n + 1)}{6} - 4\frac{n(n + 1)}{2}$ + n

= $\frac{2n}{3}$(n + 1)(2n + 1)

(ii) $2^{3} + 4^{3} + 6^{3} + 8^{3}$ + ...
The nth term will be
$a_{n}$ = a + (n - 1)d
a = 2, d = 2
$a_{n} = [2 + (n - 1)2]^{3}$
$a_{n} = (2n)^{3}$
$a_{n} = 8n^{3}$
$2^{3} + 4^{3} + 6^{3} + 8^{3}$ + ... to n terms
$\sum_{k = 2}^{n}{T_{k}} = \sum_{k = 2}^{n} 8n^{3} $

= $\sum_{k = 2}^{n} 8k^{3} $

= 8 $\sum_{k = 2}^{n} k^{3}$

= 8 $\left \{ \frac{n(n+1)}{2} \right \}^{2}$

= $2[n(n + 1)]^{2}$

iii) 1 + (1 + 2) + (1+ 2+ 3) + (1 + 2 + 3 + 4)+ ...
The nth term of the given series
= 1 + 2 + 3 + 4 + ... = $\frac{n(n + 1)}{2} = \frac{n^{2} + n}{2}$

$S_{n} = \sum_{k = 1}^{n}{T_{k}} = \sum_{k = 1}^{n} \frac{n^{2} + n }{2}$

$S_{n} = \sum_{k = 1}^{n}\frac{n^{2}}{2} + \sum_{k = 1}^{n}\frac{n}{2}$

$S_{n} = \frac{1}{2}\sum_{k = 1}^{n}n^{2}+\frac{1}{2} \sum_{k = 1}^{n} (n)$

$S_{n} = \frac{n(n + 1)(2n + 1)}{12} + \frac{n(n + 1)}{4}$
$S_{n} = \frac{n}{6}(n + 1)(n + 2)$


11th grade math

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