The polynomial P(x) is defined by

$\mathrm{P}\left(\mathrm{x}\right)={\left(\mathrm{x}+2{\mathrm{x}}^2+\dots +{\mathrm{nx}}^{\mathrm{n}}\right)}^2={\mathrm{a}}_0+{\mathrm{a}}_1\mathrm{x}+\dots +{\mathrm{a}}_{2\mathrm{n}}{\mathrm{x}}^{2\mathrm{n}}$

If ${\mathrm{a}}_{\mathrm{n}+1}+{\mathrm{a}}_{\mathrm{n}+2}+\dots +{\mathrm{a}}_{2\mathrm{n}}=\frac{\mathrm{n}(\mathrm{n}+1)\left(5{\mathrm{n}}^2+5\mathrm{n}+2\right)}{\mathrm{k}}$.

Then the value of k is

One approach is simply to expand, express the coefficients in the sum in terms of n, and simply add them. This will work, but we will opt for a slightly cleaner solution.

First note that

We now subtract off the undesired terms.
For $1\le \mathrm{k}\le \mathrm{n}$, by matching the coefficients on both sides of the definition of P, we have

${\mathrm{a}}_{\mathrm{k}}=1·(\mathrm{k}-1)+2:(\mathrm{k}-2)+\cdots +(\mathrm{k}-1)·1=\frac{1}{6}\left({\mathrm{k}}^3-\mathrm{k}\right),$

implying

${\mathrm{a}}_0+{\mathrm{a}}_1+\cdots +{\mathrm{a}}_{\mathrm{n}}=\frac{1}{6}\sum _{\mathrm{k}=1}^{\mathrm{n}}\left({\mathrm{k}}^3-\mathrm{k}\right)=\frac{1}{24}\mathrm{n}(\mathrm{n}+1)\left({\mathrm{n}}^2+\mathrm{n}-2\right)$