Question

Find the sum of $1.n^2+2(n-1{)}^2+3(n-2{)}^2+......n{.1}^2$

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

Given series is $1.n^2+2(n-1{)}^2+3(n-2{)}^2+......n{.1}^2$

General term of the series can be written as

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

Sum of the series is ${\sum }_{r=1}^{n}r.(n-r+1{)}^2$

$={\sum }_{r=1}^{n}r.\left(\right(n+1{)}^2-r{)}^2$

$={\sum }_{r=1}^{n}r.\left(\right(n+1{)}^2+r^2-2r(n+1))$

$=(n+1{)}^2{\sum }_{r=1}^{n}r+{\sum }_{r=1}^n{r}^3-2(n+1){\sum }_{r=1}^n{r}^2$

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

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

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

$=\frac{n(n+1{)}^2}{2}\left[\frac{9n+6-8n-4}{6}\right]$

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

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