Question

Show that

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

Answer 1

$\frac{1\times 2^2+2\times 3^2+\dots \dots +n(n+1{)}^2}{1^2\times 2+2^2\times 3+\dots \dots +n^2\times (n+1)}$

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

$=\frac{\sum (n^3+2n^2+n)}{\sum (n^3+n^2)}=\frac{\sum n^3+2\sum n^2+\sum n}{\sum n^3+\sum n^2}$

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

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

$=\frac{3n^2+11n+10}{3n^2+7n+2}=\frac{3n^2+6n+5n+10}{3n^2+6n+n+2}$

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

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