Question

$2\cdot 3+3\cdot 4+4\cdot 5+.....$ upto $n$ terms Find sum of $n$ of series.

Answer 1

Given series is $2.3+3.4+4.5+......n$terms

$nth$ term of the series$=(n+1)(n+2)$

sum upto $n$ terms is

${\sum }_{n=1}^n(n+1)(n+2)={\sum }_{n=1}^n{n}^2+3n+2={\sum }_{n=1}^n{n}^2+3{\sum }_{n=1}^{n}n+{\sum }_{n=1}^{n}2=\frac{n(n+1)(2n+1)}{6}+\frac{3n(n+1)}{2}+2n=\frac{n(n+1)(2n+1)+9n(n+1)+6\left(2n\right)}{6}=\frac{n\left[\right(n+1\left)\right(2n+1)+9(n+1)+12]}{6}=\frac{n[2n^2+1+3n+9n+9+12]}{6}=\frac{n[2n^2+12n+22]}{6}=\frac{n[n^2+6n+11]}{3}$