How can I calculate the following sum involving binomial terms: ∑k=0n(nk)(−1)k(k+1)2

One application of the absorption identity gets one of the factors of ${\displaystyle k+1}$ out of the denominator:
${\displaystyle \sum _{k=0}^n\left(\begin{array}{c}n\\ k\end{array}\right)\frac{{(-1)}^k}{{(k+1)}^2}=\frac{1}{n+1}\sum _{k=0}^n\left(\begin{array}{c}n+1\\ k+1\end{array}\right)\frac{{(-1)}^k}{k+1}}$
${\displaystyle =\frac{1}{n+1}\sum _{k=1}^{n+1}\left(\begin{array}{c}n+1\\ k\end{array}\right)\frac{{(-1)}^{k+1}}{k}}$
By using the basic binomial coefficient recursion formula, we can make that happen.
Let ${\displaystyle f\left(n\right)=\sum _{k=1}^n\left(\begin{array}{c}n\\ k\end{array}\right)\frac{{(-1)}^{k+1}}{k}}$ Then looking at the difference of ${\displaystyle f(n+1)}$ and ${\displaystyle f\left(n\right)}$ gives us
${\displaystyle f(n+1)-f\left(n\right)=\sum _{k=1}^{n+1}\left(\begin{array}{c}n+1\\ k\end{array}\right)\frac{{(-1)}^{k+1}}{k}-\sum _{k=1}^n\left(\begin{array}{c}n\\ k\end{array}\right)\frac{{(-1)}^{k+1}}{k}}$
${\displaystyle =\sum _{k=0}^n\left(\begin{array}{c}n\\ k\end{array}\right)\frac{{(-1)}^k}{k+1}}$
${\displaystyle =\frac{1}{n+1}\sum _{k=0}^n\left(\begin{array}{c}n+1\\ k+1\end{array}\right){(-1)}^k}$
${\displaystyle =\frac{1}{n+1}(1+\sum _{k=0}^{n+1}\left(\begin{array}{c}n+1\\ k\end{array}\right){(-1)}^{k+1})}$
${\displaystyle =\frac{1}{n+1}}$
where in the last step we used the fact that the alternating sum of the binomial coefficients is 0.
Thus
${\displaystyle f(n+1)=\sum _{k=0}^n(f(k+1)-f\left(k\right))=\sum _{k=0}^n\frac{1}{k+1}=H_{n+1}}$
Therefore,