Question

Find the sum to n terms of the series whose nth term is given by

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

Answer 1

Let $a_n$ be the nth term and '$S_n$' be the sum of the n terms of the given series.

Here $a_n=(2n-1{)}^2=4n^2-4n+1$

$\therefore S_n={\sum }_{k=1}^n{a}_k={\sum }_{k=1}^n(4k^2-4k+1)$

$=({4.1}^2-4.1+1)+({4.2}^2-4.2+1)+\dots \dots +(4n^2-4n+1)$

$=4(1^2+2^2+\dots \dots +n^2)-(1+2+\dots \dots +n)+(1+1+\dots \dots +\text{n terms})$

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

$=n[1^2+2^2+\dots \dots +n^2]-4[1+2+\dots \dots +n]+[1+1+\dots \dots +\text{n terms}]$.

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

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

$=n\left[\frac{4n^2+6n+2-6n-6+3}{3}\right]$

$=\frac{n}{3}[4n^2-1]$

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