Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11

# Find the Sum of the Series Whose Nth Term Is: (2n − 1)2 - Mathematics

Find the sum of the series whose nth term is:

(2n − 1)2

#### Solution

Let $T_n$ be the nth term of the given series.
Thus, we have:

$T_n = \left( 2n - 1 \right)^2$

Let $S_n$ be the sum of n terms of the given series.
Now,

$S_n = \sum^n_{k = 1} T_k$

$\Rightarrow S_n = \sum^n_{k = 1} \left( 2k - 1 \right)^2$

$\Rightarrow S_n = \sum^n_{k = 1} \left( 4 k^2 + 1 - 4k \right)$

$\Rightarrow S_n = {4\sum}^n_{k = 1} k^2 + \sum 1^n_{k = 1} - 4 \sum^n_{k = 1} k$

$\Rightarrow S_n = \frac{4n\left( n + 1 \right)\left( 2n + 1 \right)}{6} + n - \frac{4n\left( n + 1 \right)}{2}$

$\Rightarrow S_n = \frac{n\left( n + 1 \right)}{2}\left[ \frac{4\left( 2n + 1 \right)}{3} - 4 \right] + n$

$\Rightarrow S_n = \frac{n\left( n + 1 \right)}{2}\left( \frac{8n + 4 - 12}{3} \right) + n$

$\Rightarrow S_n = \frac{n\left( n + 1 \right)}{2}\left( \frac{8n - 8}{3} \right) + n$

$\Rightarrow S_n = 4n\left( n + 1 \right)\left( \frac{n - 1}{3} \right) + n$

$\Rightarrow S_n = \frac{n\left( 4n + 4 \right)\left( n - 1 \right) + 3n}{3}$

$\Rightarrow S_n = \frac{n}{3}\left( 4 n^2 + 4n - 4n - 4 + 3 \right)$

$\Rightarrow S_n = \frac{n}{3}\left( 4 n^2 - 1 \right)$

$\Rightarrow S_n = \frac{n}{3}\left( 2n - 1 \right)\left( 2n + 1 \right)$

Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 21 Some special series
Exercise 21.1 | Q 8.5 | Page 10