<s>[INST] Evaluate the summation $\sum_{n=1}^{10} n \left( \frac{1^2}{1 + n} + \frac{2^2}{2 + n} + ....+\frac{10^2}{10 + n}\right)$ $$P = \sum_{n=1}^{10} n \left( \frac{1^2}{1 + n} + \frac{2^2}{2 + n} + ....+\frac{10^2}{10 + n}\right)$$
My attempt: Get the constant square terms out and evaluate the summations left in $n$
$$P = 1^2\sum_{n=1}^{10}\frac{n}{1 + n} + 2^2\sum_{n=1}^{10}\frac{n}{2 + n} + ... + 10^2\sum_{n=1}^{10}\frac{n}{10 + n}$$
for the summation $\sum_{n=1}^{10}\frac{n}{k + n}$:
$$S(k) = \sum_{n=1}^{10}1-\frac{k}{k + n} \\ = 10 - k\sum_{n=1}^{10}\frac1{k+n}$$
Substituting $S(k)$ back, the sum reduces to
$$(1^2+2^2+...+10^2)10 - \sum_{k=1}^{10}\sum_{n=1}^{10}\frac{k^3}{k+n}$$
If the double summation on the right can be evaluated, the sum would be solved. Have I made a complex problem even more complex via this approach? Are there any easier ways to solve this problem?
P.S: The answer mentioned is $P = \frac{(1+2+3+....+10)^2}{2}$.
Edit: On checking with Maple, the correct answer is $55$ and not the above answer. I still do not know how maple obtained 55 (apparently $\sum n$ till 10) from this expression.
 [/INST] $P=S(10)$ where $$S(m)=\sum_{n=1}^m\sum_{k=1}^m\frac{nk^2}{k+n}=\frac{1}{2}\sum_{n=1}^m\sum_{k=1}^m\frac{nk^2+kn^2}{k+n}=\frac{1}{2}\sum_{n=1}^m\sum_{k=1}^m nk=\frac{1}{2}\left(\sum_{n=1}^m n\right)^2\color{LightGray}{=\ldots}$$
</s>