Question

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

Answer 1

$S=1\cdot 2^2+2\cdot 3^2+3\cdot 4^2+...+n{(n+1)}^2$

$S=\sum n{(n+1)}^2$

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

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

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

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

$S=n(n+1)\left(\frac{3n^2+11n+10}{12}\right)$

$S=\frac{n(n+1)(n+2)(3n+5)}{12}$