Find the sum to n terms of the series 12 + (12 + 22) + (12 + 22 + 32) + …
Question:

Find the sum to $n$ terms of the series $1^{2}+\left(1^{2}+2^{2}\right)+\left(1^{2}+2^{2}+3^{2}\right)+\ldots$

Solution:

The given series is $1^{2}+\left(1^{2}+2^{2}\right)+\left(1^{2}+2^{2}+3^{2}\right)+\ldots$

$a_{n}=\left(1^{2}+2^{2}+3^{2}+\ldots \ldots+n^{2}\right)$

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

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

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

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

$\therefore S_{n}=\sum_{k=1}^{n} a_{k}$

$=\sum_{k=1}^{n}\left(\frac{1}{3} k^{3}+\frac{1}{2} k^{2}+\frac{1}{6} k\right)$

$=\frac{1}{3} \sum_{k=1}^{n} k^{3}+\frac{1}{2} \sum_{k=1}^{n} k^{2}+\frac{1}{6} \sum_{k=1}^{n} k$

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

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

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

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

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

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

$=\frac{n(n+1)^{2}(n+2)}{12}$