Answer
$$\frac{\left( k\text{ }+\text{ }1 \right)\left( k+2 \right)\left( 2k+3 \right)}{6}$$.
Work Step by Step
The term is $\frac{k\left( k\text{ }+\text{ }1 \right)\left( 2k\text{ }+\text{ }1 \right)}{6}+{{\left( k\text{ }+\text{ }1 \right)}^{2}}$.
Taking out the $\left( k\text{ }+\text{ }1 \right)$ common term, we get,
$\begin{align}
& \frac{k\left( k\text{ }+\text{ }1 \right)\left( 2k\text{ }+\text{ }1 \right)}{6}+{{\left( k\text{ }+\text{ }1 \right)}^{2}}=\left[ \text{ }\frac{k\left( 2k\text{ }+\text{ }1 \right)}{6}+\left( k\text{ }+\text{ }1 \right) \right]\left( k\text{ }+\text{ }1 \right) \\
& =\frac{\left[ \text{ 2}{{k}^{2}}+k+6k+6 \right]}{6}\left( k\text{ }+\text{ }1 \right) \\
& =\frac{\left[ \text{ 2}{{k}^{2}}+7k+6 \right]}{6}\left( k\text{ }+\text{ }1 \right) \\
& =\frac{\left( k\text{ }+\text{ }1 \right)\left( k+2 \right)\left( 2k+3 \right)}{6}
\end{align}$
