Question

Find the sum of series to n terms:
$1\times 2+2\times 3+3\times 4+4\times 5+....$

Answer 1

$1\times 2+2\times 3+3\times 4+4\times 5+....$

In each term of sequence, first and second number forms an A.P. with common difference $1$ and $1$ respectively.

Let $T_n$ be the $n^{th}$ term of the given series, then

$T_n=\{n^{th}\text{term of}1,2,3,...\}\times \{n^{th}\text{term of}2,3,4,...\}$

$=[1+(n-1)\times 1]\times [2+(n-1)\times 1]$

$=n(n+1)$

$=n^2+n\cdots \left(1\right)$

Now, let $S_n$ be the sum of $n$ terms of the given series

We have:
$S_n=\sum T_n$

$=\sum (n^2+n)\left[\text{Using equation}\right(1\left)\right]$

$=\sum n^2+\sum n$

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

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

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

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

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

$\therefore \text{Sum of series}{S}_n=\frac{n}{3}(n+1)(n+2)$