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

$n^{\text{th}}$ term of the numerator $= n(n+1)^{2} = n^{3}+2n^{2}+n$
$n^{\text{th}}$ term of the denominator $= n^{2}(n+1) = n^{3}+n^{2}$
Let $S_N = \sum_{k=1}^{n} (k^{3}+2k^{2}+k)$ and $S_D = \sum_{k=1}^{n} (k^{3}+k^{2})$.
Using standard summation formulas:
$\sum k^{3} = \frac{n^{2}(n+1)^{2}}{4}$,$\sum k^{2} = \frac{n(n+1)(2n+1)}{6}$,$\sum k = \frac{n(n+1)}{2}$.
$S_N = \frac{n^{2}(n+1)^{2}}{4} + 2 \times \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} = \frac{n(n+1)}{12} [3n(n+1) + 4(2n+1) + 6] = \frac{n(n+1)}{12} [3n^{2}+11n+10] = \frac{n(n+1)(n+2)(3n+5)}{12}$.
$S_D = \frac{n^{2}(n+1)^{2}}{4} + \frac{n(n+1)(2n+1)}{6} = \frac{n(n+1)}{12} [3n(n+1) + 2(2n+1)] = \frac{n(n+1)}{12} [3n^{2}+7n+2] = \frac{n(n+1)(n+2)(3n+1)}{12}$.
Dividing $S_N$ by $S_D$:
$\frac{S_N}{S_D} = \frac{n(n+1)(n+2)(3n+5)}{n(n+1)(n+2)(3n+1)} = \frac{3n+5}{3n+1}$.
Thus,the result is proved.