The given sum is:
$$\sum_{n=1}^{11}n\left[\frac{1^2}{1+n}+\frac{2^2}{2+n}+\cdots+\frac{11^2}{11+n}\right]$$
I tried simplyfing it as:
$$1^2\left[\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{12}\right] + 2^2\left[\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{13}\right] +\cdots +11^2\left[\frac{1}{12}+\frac{1}{13}+\cdots+\frac{1}{22}\right]$$
Finding the sum of each diverging series seems too lengthy for a problem with obvious symmetries, so I don't think this method is correct.
$$\begin{align}S&=\sum_{n=1}^{N} \sum_{k=1}^{N} {nk^2 \over k+n}\\&=\sum_{n=1}^{N} \sum_{k=1}^{N} {nk^2 +kn^2 \over k+n}-\sum_{n=1}^{N} \sum_{k=1}^{N} {kn^2 \over k+n}\\&= \left(\sum_{n=1}^{N} \sum_{k=1}^{N} nk\right)-S \end{align}$$
Thus, $$\begin{align}S&=\frac{1}{2}\sum_{n=1}^N\sum_{k=1}^N nk\\&=\frac{1}{2}\left(\sum_{n=1}^N n\right)\left(\sum_{k=1}^N k\right)\\&=\frac{1}{2}\left(\sum_{n=1}^N n\right)^2\\&=\frac{1}{2}\left({N(N+1) \over 2}\right)^2.\end{align}$$