Let numerator
$\begin{aligned}
&S_1=1 \times 2^2+2 \times 3^2+\ldots \ldots n \times(n+1)^2 \\
&T_n=n(n+1)^2=n\left(n^2+2 n+1\right)=n^3+2 n^2+n \\
&S_1=\sum_1^n T_n=\sum_1^n n^3+2 \sum_1^n n^2+\sum_1^n n \\
&=\frac{n^2(n+1)^2}{4}+2 \frac{n(n+1)(2 n+1)}{6}+\frac{n(n+1)}{2} \\
&=\frac{n(n+1)}{12}[3 n(n+1)+4(2 n+1)+6] \\
&=\frac{n(n+1)}{12}\left[3 n^2+3 n+8 n+4+6\right] \\
&=\frac{n(n+1)}{12}\left[3 n^2+11 n+10\right]
\end{aligned}$
$=\frac{n(n+1)(3 n+5)(n+2)}{12} \quad \ldots(i)$
The denominator is denoted by $S_2$
$S_2=1^2 \times 2+2^2 \times 3+\ldots \ldots . n^2 \times(n+1)$
$n^{\text {th }}$ term $=n^2(n+1)=n^3+n^2$
$S_2=\sum_1^n n^3+\sum_1^n n^2$
$=\frac{n^2(n+1)^2}{4}+\frac{n(n+1)(2 n+1)}{6}$
$=\frac{n(n+1)}{12}\left[3 n^2+7 n+2\right]$
$=\frac{n(n+1)}{12}(3 n+1)(n+2)$
$=\frac{n(n+1)(n+2)(3 n+1)}{12} \quad \ldots(ii)$
Thus, $\quad \frac{1 \times 2^2+2 \times 3^2+\ldots \ldots \ldots+n \times(n+1)^2}{1^2 \times 2+2^2 \times 3+\ldots \ldots \ldots+n^2 \times(n+1)}=\frac{S_1}{S_2}$
$=\frac{n(n+1)(n+2)(3 n+5) / 12}{n(n+1)(n+2)(3 n+1) / 12}=\frac{3 n+5}{3 n+1}$
from (i) \& (ii)