Find the sum of `n` terms of the series `1. 2^2+2. 3^2+3. 4^2+`

Explanation

Let `T_(n)` be the nth rerm of this series, then
`T_(n)=(" nth term of "1,2,3,"...") (" nth term of "2^(2),3^(2),4^(2),".....")`
`=n(n+1)^(2)=n^(3)+2n+n`
` therefore " sum of n terms " S_(n)= sumT_(n)`
`2=sumn^(3)+2sumn^(2)+sumn`
`={(n(n+1))/(2)}+2{(n(n+1)(2n+1))/(6)}+(n(n+1))/(2)`
`=(n(n+1))/(2){(n(n+1))/(2)+(2(2n+1))/(3)+1}`
`=(n(n+1))/(12)(3n^(2)+3n+8n+4+6)`
`=(n(n+1)(3n^(2)+11n+10))/(12)=(n(n+1)(n+2)(3n+5))/(12)`