Question

Find the sum to n terms of the series whose nth term is `n^2+2^n`.

Answer 1

We have, `T_(k)=(k^(2)+2^(k))`.
`therefore S_(n)=sum_(k=1)^(n)T_(k)`
`=sum_(k=1)^(n)(k^(2)+2^(k))=sum_(k=1)^(n)k^(2)+sum_(k=1)^(n)2^(k)`
`=(1)/(6)n(n+1)(2n+1)+(2+2^(2)+2^(3)+ ...+2^(n))[because sum_(k=1)^(n)k^(2)=(1)/(6)n(n+1)(2n+1)]`
`=(1)/(6)n(n+1)(2n+1)+(2(2^(n)-1))/((2-1)) " "[because 2+2^(2)+ ... +2^(n)" is a GP"]`
`=(1)/(6)n(n+1)(2n+1)+2(2^(n)-1).`
Hence, the required sum is `(1)/(6)n(n+1)(2n+1)+2(2^(n)-1).`