Find the sum of series upto $n$ terms whose $n^{t h}$ term is ${(2 n - 1)}^{2}$

Open in App

Solution

Consider the problem

Given
$\begin{matrix} a_{n} = {(2 n - 1)}^{2} = {(2 n)}^{2} + {(1)}^{2} - 2 (2 n) (1) = 4 n^{2} + 1 - 4 n = 4 n^{2} - 4 n + 1 \end{matrix}$

Sum of $n$ term is
$\begin{matrix} s_{n} = \sum_{n = 1}^{n} a_{n} = \sum_{n = 1}^{n} 4 n^{2} - 4 n + 1 = \sum_{n = 1}^{n} 4 n^{2} - \sum_{n = 1}^{n} 4 n + \sum_{n = 1}^{n} 1 = 4 \sum_{n = 1}^{n} n^{2} - 4 \sum_{n = 1}^{n} n + \sum_{n = 1}^{n} 1 = 4 (\frac{n (n + 1) (2 n + 1)}{6}) - 4 (\frac{n (n + 1)}{2}) + n = n (\frac{n (n + 1) (2 n + 1)}{6}) - 4 (\frac{n (n + 1)}{2} + 1) = n (\frac{2}{3} (n + 1) (2 n + 1) - 2 (n + 1) + 1) = n (\frac{2 (n + 1) (2 n + 1) - 6 (n + 1) + 3}{3}) = \frac{n}{3} [(2 n + 2) (2 n + 1) - 6 n - 6 + 3] = \frac{n}{3} [4 n^{2} + 2 n + 4 n + 2 - 6 n - 3] = \frac{n}{3} [4 n^{2} + 6 n - 6 n - 1] = \frac{n}{3} [4 n^{2} - 1] = \frac{n}{3} [{(2 n)}^{2} - {(1)}^{2}] = \frac{n}{3} [(2 n - 1) (2 n + 1)] \end{matrix}$

Hence, the required sum is $\frac{n}{3} [(2 n - 1) (2 n + 1)]$

Suggest Corrections

Similar questions

n

n t h

(2 n - 1)^{2}

Find the sum to n terms of the series whose n^th terms is given by (2n – 1)²

n

n^{t h}

{(2 n - 1)}^{2}

n

n^{t h}

n^{3} + \frac{3}{2} n

n

n t h

n^{2} + 2^{n}

Join BYJU'S Learning Program