Question

Find the sum to n terms

$3\times 1^2+5\times 2^2+7\times 3^2+\dots \dots$

Answer 1

The given series is

$3\times 1^2+5\times 2^2+7\times 3^2+\dots \dots +$ to n terms.

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

$\therefore a_n=$ [nth term of 3, 5, 7...] $[\text{nth term of 1, 2, 3, ...}{]}^2$

$=(2n+1)(n{)}^2=2n^3+n^2$

$\therefore S_n={\sum }_{k=1}^n{a}_k={\sum }_{k=1}^n(2k^3+k^2)$

$=({2.1}^3+1^2)+({2.2}^3+2^2)+({2.3}^3+3^2)+\dots \dots +(2.n^3+n^2)$

$=2(1^3+2^3+3^3+\dots \dots +n^3)+(1^2+2^2+3^2+\dots \dots +n^2)$

$=2{\left[\frac{n(n+1)}{2}\right]}^2+\frac{n(n+1)(2n+1)}{6}$

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

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

$=\frac{n(n+1)}{2}\left[\frac{3(n^2+n)+(2n+1)}{3}\right]$

$=\frac{n(n+1)}{2}\left[\frac{3n^2+3n+2n+1}{3}\right]$

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

$=\frac{n(n+1)(3n^2+5n+1)}{6}$