Solution

$$\displaystyle { a }_{ n }={ n }^{ 2 }+{ 2 }^{ n }\\ \therefore { S }_{ n }=\sum _{ k=1 }^{ n }({ { k }^{ 2 } } +2k)=\sum _{ k=1 }^{ n }{ { k }^{ 2 }+\sum _{ k=1 }^{ n }{ { 2 }^{ k } } } ........ (1)$$
Consider $$ \sum _{ k=1 }^{ n }{ { 2 }^{ k } } ={ 2 }^{ 1 }+{ 2 }^{ 2 }+{ 2 }^{ 3 }\quad +\quad ...$$
The above series $$ 2, { 2 }^{ 2 }, { 2 }^{ 3 }, ...$$ is a G.P. with both the first term and common ratio equal to 2.
$$\displaystyle \therefore \sum _{ k=1 }^{ n }{ { 2 }^{ k } } =\frac { { (2) }\left[ ({ 2 })^{ n }-1 \right] }{ 2-1 } =2({ 2 }^{ n }-1)\quad \quad \quad \quad (2)$$
Therefore, from (1) and (2), we obtain
$$\displaystyle { S }_{ n }=\sum _{ k=1 }^{ n }{ { k }^{ 2 } } +2({ 2 }^{ n }-1)=\frac { n(n+1)(2n+1) }{ 6 } +2({ 2 }^{ n }-1)$$