Question

If $\frac{1\times 2^2+2\times 3^2+3\times 4^2+\cdots +n(n+1{)}^2}{1^2\times 2+2^2\times 3+3^2\times 4+\cdots +n^2(n+1)}=\frac{8}{7}$, then $n=$

Answer 1

$\frac{1\times 2^2+2\times 3^2+3\times 4^2+\cdots +n(n+1{)}^2}{1^2\times 2+2^2\times 3+3^2\times 4+\cdots +n^2(n+1)}$
Let $S_1=1\times 2^2+2\times 3^2+3\times 4^2+\cdots +n(n+1{)}^2$
$T_n=n(n+1{)}^2$ which is $n^{\text{th}}$ term of $S_1$
So $S_1=\sum (n^3+2n^2+n)$
Let $S_2=1^2\times 2+2^2\times 3+3^2\times 4+\cdots +n^2(n+1)$
$T_n=n^2(n+1)$ which is $n^{\text{th}}$ term of $S_2$
So $S_2=\sum (n^3+n^2)$
$\therefore \frac{1\times 2^2+2\times 3^2+3\times 4^2+\cdots +n(n+1{)}^2}{1^2\times 2+2^2\times 3+3^2\times 4+\cdots +n^2(n+1)}=\frac{\sum (n^3+2n^2+n)}{\sum (n^3+n^2)}=\frac{\frac{1}{4}{n}^2(n+1{)}^2+\frac{2}{6}n(n+1\left)\right(2n+1)+\frac{1}{2}n(n+1)}{\frac{1}{4}{n}^2(n+1{)}^2+\frac{1}{6}n(n+1\left)\right(2n+1)}=\frac{3n(n+1)+4(2n+1)+6}{3n(n+1)+2(2n+1)}=\frac{3n^2+11n+10}{3n^2+7n+2}=\frac{(3n+5)(n+2)}{(3n+1)(n+2)}=\frac{3n+5}{3n+1}=1+\frac{4}{3n+1}$

Now,
$1+\frac{4}{3n+1}=\frac{8}{7}\Rightarrow n=9$