Find the sum of n terms of the series whose nth term is: `n(n+3)`

Explanation

Given

`a_n=n(n+3)`

`a_n=n^2+3n`


The sum of `n` terms is

`S_n=sum_(n=1)^n a_n`

`S_n=sum_(n=1)^n (n^2+3n)`

`S_n=sum_(n=1)^n n^2+sum_(n=1)^n 3n`

`S_n=sum_(n=1)^n n^2+3sum_(n=1)^n n`

`S_n=(n(n+1)(2n+1))/6+3(n(n+1))/2`

`S_n=(n(n+1))/2(((2n+1))/3+3)`

`S_n=(n(n+1))/2(((2n+1)+9)/3)`

`S_n=(n(n+1))/2((2n+10)/3)`

`S_n=(n(n+1))/(2)xx(2(n+5))/3`

`S_n=(n(n+1)(n+5))/3`

Thus, the required sum is `(n(n+1)(n+5))/3`