Answer
$126$
Work Step by Step
We calculate the given sum by expanding it: we calculate each term for $k=3,4,5,6$ and add them:
$$\begin{align*}
\sum_{k=3}^6 (k+1)^2&=(3+1)^2+(4+1)^2+(5+1)^2+(6+1)^2\\
&=4^2+5^2+6^2+7^2\\
&=16+25+36+49\\
&=126.
\end{align*}$$
