Question

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

• A. $\frac{n\left(n+1\right)^{2}\left(n+2\right)}{12}$
• B. $\frac{n\left(n+1\right)^{2}\left(n+3\right)}{12}$
• C. $\frac{n\left(n+2\right)^{2}\left(n+1\right)}{12}$
• D. $\frac{n\left(n+2\right)^{2}\left(n+1\right)}{14}$

Answer 1

Answer: (A)

$T_{n} = \left(1^{2}+2^{2}+3^{2} +...+n^{2}\right)$

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

$ = \frac{1}{6}\left[2n^{3}+3n^{2} +n\right] $

$ \therefore S_{n} = \sum\limits_{k=1}^{n} T_{k} $

$= \frac{1}{6}\left[2\sum\limits _{k=1}^{n} k^{3} +3\sum\limits _{k=1}^{n} k^{2} +\sum\limits _{k=1}^{n} k\right]$

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

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

$= \frac{n\left(n+1\right)}{12}\left[\frac{n\left(n+1\right)+2n+1+1}{1}\right]$

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