Find the sum to `n` terms of the series `1xx2xx3+2xx3xx4+3xx4xx5+dot`

Answer

`(n(n+1)(n+2)(n+3))/(4)`

Explanation

`T_(n)=n(n+1)(n+2)`
Let `S_(n)` denote the sum to n terms of the given series. Then,
`S_(n)=sum_(k=1)^(n)T_(k)=sum_(k=1)^(n)k(k+1)(k+2)`
`=sum_(k=1)^(n)(k^(3)+3k^(2)+2k)`
`=(sum_(k=1)^(n)k^(3))+3(sum_(k=1)^(n)k^(2))+2(sum_(k=1)^(n)k)`
`=((n(n+1))/2)^(2)+(3n(n+1)(2n+1))/6+(2n(n+1))/2`
`=(n(n+1))/2{(n(n+1))/2+(2n+1)+2}`
`=(n(n+1))/4{n^(2)+n+4n+2+4}`
`=(n(n+1))/4(n^(4)+5n+6)`
`=(n(n+1)(n+2)(n+3))/4`