Question

Find the sum of $1\cdot 2\cdot 3+2\cdot 3\cdot 4+3\cdot 4\cdot 5+\cdots$ to $n$ terms.

Answer 1

$S=1\cdot 2\cdot 3+2\cdot 3\cdot 4+3\cdot 4\cdot 5+...+n\cdot (n+1)\cdot (n+2)$

$S=\sum n\cdot (n+1)\cdot (n+2)$

$S=\sum n^3+3n^2+2n$

$S=\sum n^3+3\sum n^2+2\sum n$

Sum of cube of $n$ natural number will be ${\left(\frac{n(n+1)}{2}\right)}^2$

Sum of square of $n$ natural number will be $\left(\frac{n(n+1)(2n+1)}{6}\right)$

Sum of $n$ natural number will be $\left(\frac{n(n+1)}{2}\right)$

$S={\left(\frac{n(n+1)}{2}\right)}^2+3\left(\frac{n(n+1)(2n+1)}{6}\right)+2\left(\frac{n(n+1)}{2}\right)$

$S=n(n+1)[\left(\frac{n(n+1)}{4}\right)+\frac{2n+1}{2}+1]$

$S=n(n+1)\left[\frac{n^2+n+4n+2+4}{4}\right]$

$S=n(n+1)\left(\frac{n^2+5n+6}{4}\right)$

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