Question

Find m if the following equation holds true $1.3+2.4+3.5+...+n(n+2)=\frac{1}{m}n(n+1)(2n+7)$

Answer 1

$n$th term of the given series is, $a_n=n(n+2)$

$a_n=n^2+2n$

Sum of $n$ terms is $S_n={\sum }_{k=1}^n{k}^2+2{\sum }_{k=1}^{n}k$

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

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

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

$=n(n+1)\left[\frac{2n+7}{6}\right]$

$=\frac{n(n+1)(2n+7)}{6}$

$\Rightarrow \frac{1}{m}=\frac{1}{6}$

$m=6$