the series is like : `1(1+1)(1+2) , 2(2+1)(2+2)`
so the nth term is `a_n= n(n+1)(n+2)`
`S_n = sum a_n = sum(n)(n^2 + 3n+2)`
`= sum(n^3 + 3n^2 +2n)`
`= sum n^3 + 3 sum n^2 + 2 sum n`
`= (n^2(n+1)^2)/4 + (3(n)(n+1)(2n+1))/6 + (2n(n+1))/2`
`= (n(n+1))/4*[n(n+1) + 2(2n+1) + 4]`
`= (n(n+1))/4[n^2 + 5n +6]`
`S_n= (n(n+1)(n+2)(n+3))/4`
answer
