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

1.	Find the sum of n terms of the series whose nth term is: `n(n+3)`
Answer» We have , `t_(k)=k(k+3)=(k^(2)+3k)`. `therefore ` sum to n terms is given by `S_(n)=sum_(k=1)^(n)T_(k)` `=sum_(k=1)^(n)(k^(2)+3k)=sum_(k=1)^(n)k^(2)+3sum_(k=1)^(n)k` `=(1)/(6)n(n+1)(2n+1)+3(1)/(2)n(n+1)=(1)/(6){n(n+1)(2n+1)+9n(n+1)}` `=(1)/(6)n(n+1)(2n+1+9)=(1)/(6)n(n+1)2(n+5)=(1)/(3)n(n+1)(n+5).` Hence, the required sum is `(1)/(3)n(n+1)(n+5).`

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