Show that `S_(n)=(n(2n^(2)+9n+13))/(24)`.

Explanation

Let `T_(n)` and `T_(n)'` be the nth terms of the series in numerator and denominaror of LHS. Then,
`therefore T_(n)=n(n+1)^(2)" and " T_(n)'=n^(2)(n+1)`
`therefore LHS=(sumT_(n))/(sumT_(n))=(sumn(n+1)^(2))/(sumn^(2)(n+1))=(sum(n^(3)+2n^(2)+n))/(sum(n^(3)+n^(2)))`
`=(sumn^(3)+2sumn^(2)+sumn)/(sumn^(3)+sumn^(2))`
`=({n(n+1)/(2)}^(2)+2{n(n+1)(2n+1)/(6)}+{n(n+1)/(2)})/({n(n+1)/(2)}^(2)+{(n(+1)(2n+1))/(6)})`
`=((n(n+1))/(2){(n(n+1))/(2)+(2(2n+1))/(3)+1})/((n(n+1))/(2){(n(n+1))/(2)+(2n+1)/(3)})`
`=((1)/(6)(3n^(2)+3n+8n+4+6))/((1)/(6)(3n^(2)+3n+4n+2))`
`=((3n^(2)+11n+10))/((3n^(2)+7n+2))=((3n+5)(n+2))/((3n+1)(n+2))=((3n+5))/((3n+1))=RHS`.