Question

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

Answer 1

The $n^{th}$ term of the numerator $=n{(n+1)}^2=n^2+{2n}^2+n$
And $n^{th}$ term of the denominator $=n^2(n+1)=n^3+n^2$
$\therefore \frac{1\times 2^2+2\times 3^2+...+n\times {(n+1)}^2}{1^2\times 2+2^2\times 3+...+n^2\times (n+1)}=\frac{{\sum }_{k-1}^n{a}_k}{{\sum }_{k-1}^n{a}_k}=\frac{{\sum }_{k-1}^n(k^3+k^2+k)}{{\sum }_{k-1}^n(k^3+k^2)}...\left(1\right)$
Here, $\sum _{k-1}^n(k^3+2k^2+k)=\frac{n^2{(n+1)}^2}{4}+\frac{2n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}=\frac{n(n+1)}{2}[\frac{n(n+1)}{2}+\frac{2}{3}(2n+1)+1]=\frac{n(n+1)}{2}\left[\frac{{3n}^2+3n+8n+4+6}{6}\right]=\frac{n(n+1)}{12}[{3n}^2+11n+10]=\frac{n(n+1)}{12}[{3n}^2+6n+5n+10]=\frac{n(n+1)}{12}\left[3n\right(n+2)+5(n+2\left)\right]=\frac{n(n+1)(n+2)(3n+5)}{12}...\left(2\right)\text{Also},{\sum }_{k-1}^n(k^3+k^2)=\frac{n^2{(n+1)}^2}{4}+\frac{n(n+1)(2n+1)}{6}=\frac{n(n+1)}{2}[\frac{n(n+1)}{2}+\frac{2n+1}{3}]=\frac{n(n+1)}{2}\left[\frac{{3n}^2+3n+4n+2}{6}\right]=\frac{n(n+1)}{12}[{3n}^2+7n+2]=\frac{n(n+1)}{12}[{3n}^2+6n+n+2]=\frac{n(n+1)}{12}\left[3n(n+2)+1(n+2)\right]=\frac{n(n+1)(n+2)(3n+1)}{12}...\left(3\right)$
From (1), (2) and (3) we obtain
$\frac{1\times 2^2+2\times 3^2+...+n\times {(n+1)}^2}{1^2\times 2+2^2\times 3+...+n^2\times (n+1)}=\frac{\frac{n(n+1)(n+2)(3n+5)}{12}}{\frac{n(n+1)(n+2)(3n+1)}{12}}=\frac{n(n+1)(n+2)(3n+5)}{n(n+1)(n+2)(3n+1)}=\frac{3n+5}{3n+1}$
Thus, the given result is proved.