Question

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

Answer 1

$1^{st}$ term $1\times 2$

$2^{nd}$ term $2\times 3$

$3^{rd}$ term $3\times 4$

$n^{th}$ term $n\times (n+1)$

Step 1 : Find $n^{th}$ team (an)

Here, an $=n(n+1)$

$=n^n+n$

step $2:S_n=\sum _{n=1}^{n}an$

$=\sum _n^{n=1}{n}^2+n$

$=\sum _{n=1}^n{n}^2+\sum _{n=1}^{n}n$

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

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

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

$=\frac{n(n+1)}{2}\left(\frac{2n+4}{3}\right)$

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