Question

Find m if the following equation holds true: $1.2+4.3+9.4+...n^2(n+1)=\frac{n}{m}(n+1)(n+2)(3n+1)$

Answer 1

$n$th term, $a_n=n^2(n+1)=n^3+n^2$

Sum of first $n$ terms,

$S_n={\sum }_{n=1}^n{a}_n$

$S_n={\sum }_{n=1}^n{n}^3+{\sum }_{n=1}^n{n}^2$

$=\frac{n^2(n+1{)}^2}{4}+\frac{n(n+1)(2n+1)}{6}$

$=n(n+1)\{\frac{n(n+1)}{4}+\frac{2n+1}{6}\}$

$=n(n+1)\left\{\frac{3n^2+7n+2}{12}\right\}$

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

So, $m=12$