Find the sum to n terms of the series `1^(2)+(1^(2)+2^(2))+(1^(2)+2^(2)+3^(2))+.....`

Explanation

Let `T_(n)` denote the `n^(th)` term of the given series.
`thereforeT_(n)=1^(2)+2^(2)+3^(2)+.....n^(2)=sumn^(2)`
`=(n(n+1)(2n+1))/(6)=(1)/(6)n(2n^(2)+3n+1)=(2n^(3)+3n^(2)+n)/(6)=(1)/(3)n^(3)+(1)/(2)n^(2)+(1)/(6)n`
`thereforeS_(n)=sumT_(n)=(1)/(3)sumn^(3)+(1)/(2)sumn^(2)+(1)/(6)sumn`
`thereforeS_(n)=sumT_(n)=(1)/(3)sumn^(3)+(1)/(2)sumn^(2)+(1)/(6)sumn`
`=(1)/(3)(n^(2)(n+1)^(2))/(4)+(1)/(2)(n(n+1)(2n+1))/(6)+(1)/(6)(n(n+1))/(2)`
`=(n(n+1))/(12)[n(n+1)+(2n+1)+1]=(n(n+1))/(12)[n^(2)+3n+2]`
`=(n(n+1)(n+1)(2n+1))/(12)=(n(n+1)^(2)(n+2))/(12)`