Express the Sum in Terms of N $\sum _ { k = 1 } ^ { n } \left( k ^ { 2 } + 5 k + 5 \right)$

Express the sum in terms of n. $\sum _ { k = 1 } ^ { n } \left( k ^ { 2 } + 5 k + 5 \right)$ (Hint: Use the theorem on sums to write the sum as $\sum _ { k = 1 } ^ { n } k ^ { 2 } + 5 \sum _ { k = 1 } ^ { n } k + \sum _ { k = 1 } ^ { n } 5$ . Next, use the following formulas $$\begin{array} { l } 1 ^ { 2 } + 2 ^ { 2 } + 3 ^ { 2 } + \ldots + n ^ { 2 } = \frac { n ( n + 1 ) ( 2 n + 1 ) } { 6 } \\1 + 2 + 3 + \ldots + n = \frac { n ( n + 1 ) } { 2 }\end{array}$$ and the theorem on the sum of a constant.)

A) $\frac { n ^ { 3 } - 9 n ^ { 2 } + 23 n } { 3 }$
B) $\frac { n ^ { 3 } + 9 n ^ { 2 } - 23 n } { 3 }$
C) $$\frac { n ^ { 3 } + 9 n ^ { 2 } + 23 n } { 3 }$$
D) $$\frac { n ^ { 3 } - 9 n ^ { 2 } - 23 n } { 3 }$$
E) $\frac { n ^ { 3 } + 3 n ^ { 2 } + 46 n } { 3 }$