Question

The sum of the series $1\times n+2(n-1)+3(n-2)+\dots \dots +(n-1)\times 2+n\times 1$ is

• A. $\frac{n(n+1)(n+2)}{6}$
• B. $\frac{n(2n+1)(n+1)}{4}$
• C. $\frac{n(n+1)(n+2)(n+3)}{6}$
• D. $\frac{n(2n+1)(n+1)(n+2}{4}$

Answer 1

The correct option is A $\frac{n(n+1)(n+2)}{6}$

Let $T_r$ be the $r^{th}$ term of the given series. Then,
$T_r=r\times \{n-(r-1\left)\right\}$
$=r(n-r+1)$
$=r\left\{\right(n+1)-r\}$
$=(n+1)r-r^2$
$\therefore S=\sum _{r=1}^n{T}_r=\sum _{r=1}^n\left[\right(n+1)r-r^2]$
$=(n+1)\left(\sum _{r=1}^{n}r\right)-\left(\sum _{r=1}^n{r}^2\right)$
$=(n+1)\frac{n(n+1)}{2}-\frac{n(n+1)(2n+1)}{6}$
$=\frac{n(n+1)(n+2)}{6}$

Alternate solution:
When $n=1$, then
$S=1$
Now, checking the options,
$\frac{n(n+1)(n+2)}{6}=\frac{1\left(2\right)\left(3\right)}{6}=1$
$\frac{n(2n+1)(n+1)}{4}=\frac{1\left(3\right)\left(2\right)}{4}=\frac{3}{2}$
$\frac{n(n+1)(n+2)(n+3)}{6}=\frac{1\left(2\right)\left(3\right)\left(4\right)}{6}=4$
$\frac{n(2n+1)(n+1)(n+2}{4}=\frac{1\left(3\right)\left(2\right)\left(3\right)}{4}=\frac{9}{2}$

Hence, the correct option is $\frac{n(n+1)(n+2)}{6}$