Show that
Question:

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

Solution:

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

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

$\frac{1 \times 2^{2}+2 \times 3^{2}+\ldots .+n \times(n+1)^{2}}{1^{2} \times 2+2^{2} \times 3+\ldots .+n^{2} \times(n+1)}=\frac{\sum_{K=1}^{n} a_{K}}{\sum_{K=1}^{n} a_{K}}=\frac{\sum_{K=1}^{n}\left(K^{3}+2 K^{2}+K\right)}{\sum_{K=1}^{n}\left(K^{3}+K^{2}\right)}$ (1)

Here, $\sum_{K=1}^{n}\left(K^{3}+2 K^{2}+K\right)$

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

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

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

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

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

$=\frac{\mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2)(3 \mathrm{n}+5)}{12}$ (2)

Also, $\sum_{K=1}^{n}\left(K^{3}+K^{2}\right)=\frac{n^{2}(n+1)^{2}}{4}+\frac{n(n+1)(2 n+1)}{6}$

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

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

$=\frac{\mathrm{n}(\mathrm{n}+1)}{12}\left[3 \mathrm{n}^{2}+7 \mathrm{n}+2\right]$

$=\frac{\mathrm{n}(\mathrm{n}+1)}{12}\left[3 \mathrm{n}^{2}+6 \mathrm{n}+\mathrm{n}+2\right]$

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

$=\frac{n(n+1)(n+2)(3 n+1)}{12}$ (3)

From (1), (2), and (3), we obtain

$\frac{1 \times 2^{2}+2 \times 3^{2}+\ldots+n \times(n+1)^{2}}{1^{2} \times 2+2^{2} \times 3+\ldots+n^{2} \times(n+1)}=\frac{\frac{n(n+1)(n+2)(3 n+5)}{12}}{\frac{n(n+1)(n+2)(3 n+1)}{12}}$

$=\frac{n(n+1)(n+2)(3 n+5)}{n(n+1)(n+2)(3 n+1)}=\frac{3 n+5}{3 n+1}$

Thus, the given result is proved.