Given series is $\large\frac{1\times2^2+2\times 3^2+........n\times (n+1)^2}{1^2\times 2+2^2\times 3+.........n^2\times (n+1)}$
Step 1
The numerator is $1\times2^2+2\times 3^2+........n\times (n+1)^2$
$n^{th}$ term of the numerator is $t_n=n(n+1)^2=n^3+2n^2+n$
We know that sum of $n$ terms of any series $=\sum t_n$
$\therefore $ Numerator$=\sum t_n=\sum(n^3+2n^2+n)$
$\qquad \qquad\:=\sum n^3+2.\sum n^2+\sum n$
We know that $\sum n^3=\large\frac{n^2(n+1)^2}{4}$, $\sum n^2=\large\frac{n(n+1)(2n+1)}{6}$ and $\sum n=\large\frac{n(n+1)}{2}$
$\Rightarrow\: $ Numerator $=\large\frac{n^2(n+1)^2}{4}$$+2.\large\frac{n(n+1)(2n+1)}{6}$$+\large\frac{n(n+1)}{2}$
$\qquad\:=\large\frac{n(n+1)}{2}$$\bigg[\large\frac{n(n+1)}{2}$$+\large\frac{2(2n+1)}{3}$$+1\bigg]$
$\qquad\:=\large\frac{n(n+1)}{2}$$\bigg[\large\frac{3n^2+3n+8n+4+6}{6}\bigg]$
$\qquad\:=\large\frac{n(n+1)}{2}$$\bigg[\large\frac{3n^2+11n+10}{6}\bigg]$
$\qquad\:=\large\frac{n(n+1)}{12}$$(3n^2+6n+5n+10)$$=\large\frac{n(n+1)}{12}$$(3n+5)(n+2)$
$\qquad\:=\large\frac{n(n+1)(n+2)(3n+5)}{12}$............(i)
Step 2
Denominator is $1^2\times 2+2^2\times 3+.........n^2\times (n+1)$
$n^{th}$ term of this series is $t_n=n^2(n+1)=n^3+n^2$
We know that sum of $n$ terms of this series is $\sum t_n$
$\therefore$ Denominator $=\sum t_n=\sum( n^3+n^2)$
$\qquad\:=\sum n^3+\sum n^2$
$\qquad \:=\large\frac{n^2(n+1)^2}{4}+\frac{n(n+1)(2n+1)}{6}$
$\qquad\:=\large\frac{n(n+1)}{2}$$\bigg[\large\frac{n(n+1)}{2}+\frac{2n+1}{3}\bigg]$
$\qquad\:=\large\frac{n(n+1)}{2}$$\bigg[\large\frac{3n^2+3n+4n+2}{6}\bigg]$
$\qquad\:=\large\frac{n(n+1)}{2}$$\bigg[\large\frac{3n^2+7n+2}{6}\bigg]$
$\qquad\:=\large\frac{n(n+1)(3n^2+6n+n+2)}{12}$
$\qquad\:\large\frac{n(n+1)(3n+1)(n+2)}{12}$...........(ii)
Step 3
$\therefore$ The given series $=\large\frac{Numerator}{Denominator}$
$\qquad\: = \large\frac{n(n+1)(n+2)(3n+5)}{n(n+1)(3n+1)(n+2)}$
$\qquad\:=\large\frac{3n+5}{3n+1}$
Hence proved.