Show that : $\frac{1\times 2^2+2\times 3^2+...+n\times (n+1{)}^2}{1^2\times 2+2^2\times 3+...+n^2\times (n+1)}=\frac{3n+5}{3n+1}$

Step $1$. Solving given data.

$\because n^{th}$ term of the numerator $=n(n+1{)}^2=n^3+2n^2+n$

$n^{th}$ term of the denominator $=n^2(n+1)=n^3+n^2$

$\text{L.H.S}=\frac{1\times 2^2+2\times 3^2+...+n\times (n+1{)}^2}{1^2\times 2+2^2\times 3+...+n^2\times (n+1)}=\frac{{\sum }_{n=1}^n{a}_n}{{\sum }_{n=1}^n{a}_n}$

$\text{L.H.S}=\frac{{\sum }_{n=1}^n(n^3+2n^2+n)}{{\sum }_{n=1}^n(n^3+n^2)}.........\left(i\right)$

Step $2$. Solving numerator of $\text{L.H.S.}$ separately for $n^{th}$ term.

Now, numerator,
${\sum }_{n=1}^n(n^3+2n^2+n)$

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

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

$=\frac{n(n+1)}{2}\left[\frac{3n^2+3n+8n+4+6}{6}\right]$

$=\frac{n(n+1)}{12}[3n^2+11n+10]$

$=\frac{n(n+1)}{12}[3n^2+6n+5n+10]$

$=\frac{n(n+1)}{12}\left[3n\right(n+2)+5(n+2\left)\right]$

Numerator $=\frac{n(n+1)(n+2)(3n+5)}{12}.......\left(ii\right)$

Step $3$. Solving denominator of $\text{L.H.S.}$ separately for $n^{th}$ term.

Now denominator,

${\sum }_{n=1}^n(n^3+n^2)=\frac{n^2(n+1{)}^2}{4}+\frac{n(n+1)(2n+1)}{6}$

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

$=\frac{n(n+1)}{2}\left[\frac{3n^2+3n+4n+2}{6}\right]$

$=\frac{n(n+1)}{2}\left[\frac{3n^2+7n+2}{6}\right]$
$=\frac{n(n+1)}{2}\left[\frac{3n^2+6n+n+2}{6}\right]$

$=\frac{n(n+1)}{2}\left[\frac{3n(n+2+1(n+2)}{6}\right]$

Denominator $=\frac{n(n+1)(3n+1)(n+2)}{12}.......\left(iii\right)$

Step $4$. Solve for prove.

Putting $\left(ii\right)$ and $\left(iii\right)$ in $\left(i\right)$

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

$\text{L. H.S}=\frac{(3n+5)}{(3n+1)}$

$\text{L.H.S = R.H.S}$

Hence proved.