Question

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

Answer 1

$\text{The given series is},1\times 2\times 3+2\times 3\times 4+3\times 4\times 5+...\text{nth term},a_n=n(n+1)(n+2)=(n^2+n)(n+2)=n^3+{3n}^2+2n\therefore S_n={\sum }_{k=1}^n{a}_k={\sum }_{k=1}^n{k}^3+3{\sum }_{k=1}^n{k}^2+2{\sum }_{k=1}^{n}k={\left[\frac{n(n+1)}{2}\right]}^2+\frac{3n(n+1)(2n+1)}{6}+n(n+1)={\left[\frac{n(n+1)}{2}\right]}^2+\frac{n(n+1)(2n+1)}{2}+n(n+1)=\frac{n(n+1)}{2}[\frac{n(n+1)}{2}+2n+1+2]=\frac{n(n+1)}{2}\left[\frac{n^2+n+4n+6}{2}\right]=\frac{n(n+1)}{4}(n^2+5n+6)=\frac{n(n+1)}{4}(n^2+2n+3n+6)=\frac{n(n+1)\left[n\right(n+2)+3(n+2\left)\right]}{4}=\frac{n(n+1)(n+2)(n+3)}{4}$