Question

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

Answer 1

Let $T_n$ be the nth term of this series. Then, $T_n$= (nth term of 1,2,3...)$\times$ (nth term of 2, 3, 4...) $=[1+(n-1)\times 1].[2+(n-1)\times 1]=[1+n-1].[2+n-1]=n(n+1)=n^2+nLet{S}_n\text{be the sum to n terms of the given series; Then,}{S}_n={\sum }_{n-1}^n{T}_n={\sum }_{n-1}^n(n^2+n)={\sum }_{n-1}^n{n}^2+{\sum }_{n-1}^{n}n=\frac{n(n+1)(2n+1)}{6}+\frac{(n+1)n}{2}=\frac{n(n+1)(2n+1)+3n(n+1)}{6}=\frac{n(n+1)[2n+1+3]}{6}=\frac{n(n+1)[2n+4]}{6}=\frac{n(n+1)\times 2(n+2)}{6}=\frac{2n}{6}(n+1)(n+2)Hence,S_n=\frac{n}{3}(n+1)(n+2)$