Find the sum to n terms of the series, whose `n^(t h)`terms is given by : `n(n+1)(n+4)`

Answer

N/a

Explanation

nth term of the series
`= n(n + 1)(n + 4)`
`= n(n^(2) + 4n + n + 4) = n(n^(2) + 5n + 4)`
`= n^(3) + 5n^(2) + 4n`
`rArr` `S_(n) = sum(n^(3) + 5n^(2) + 4n)`
`= sum n^(3) + 5 sum n^(2) + 4 sum n`
` = (n^(2)(n + 1)^(2))/(4) + (5n(n + 1)(2n + 1))/(6) + (4n(n + 1))/(2)`
`= (n(n + 1))/(2) [(n(n + 1))/(2) + (5(2n + 1))/(3) + 4]`
`= (n(n + 1))/(2) [(3n^(2) + 3n + 20n + 10 + 24)/(6)]`
`= (n(n + 1))/(2) * (3n^(2) + 23n + 34)/(6)`
`= (n(n + 1))/(12)[3n^(2) + 17n + 6n + 34]`
`= (n(n + 1))/(12) [n(3n + 17) + 2(3n + 17)]`
`= (n(n + 1)(3n + 17)(n + 2))/(12)`