For every integer $n\ge1$ find the minimum value of $W_n(x)=x^{2n}+2x^{2n-1}+3x^{2n-2}+\dots+2nx$ 
Let $$W_n(x)=x^{2n}+2x^{2n-1}+3x^{2n-2}+\dots+2nx$$ For every integer $n\ge1$ find the minimum value of the given polynomial.

My approach was to rewrite $W_n(x)$ as
$$W_n(x)=(x^{2n}-1)+(2x^{2n-1}+2)+(3x^{2n-2}-3)+\dots+(2nx+2n)-n$$ After that I experimented a bit and fully convinced myself that $(-n)$ is the minimum value of the given polynomial, however I don't how to prove my claim.
How do I proceed? Any help appreciated.
 A: It suffices to verify the identity
$$
W_n(x)+n
=
(x+1)^2
\sum_{k=1}^n kx^{2(n-k)}
$$
To achieve the verification, simply expand the RHS and compare like powers . . .

It's easily seen that the leading terms are the same, and the constant terms are the same.

Next, fix $m$ with $1\le m\le 2n-1$.

For the LHS, the coefficient of $x^m$ is $2n+1-m$.

For the RHS, consider consider two cases according as $m$ is even or odd.

First suppose $m$ is even.

Then $m=2(n-k)$, with $1\le k\le n-1$, so the $x^m$ term in the expansion of the RHS is
\begin{align*}
&
x^2\Bigl((k+1)x^{2(n-k)-2}\Bigr)+kx^{2(n-k)}
\\[4pt]
=\,&
(2k+1)x^{2(n-k)}
\\[4pt]
=\,&
(2n+1-m)x^m
\end{align*}
Next suppose $m$ is odd.

Then $m=2(n-k)+1$, with $1\le k\le n$, so the $x^m$ term in the expansion of the RHS is
\begin{align*}
&
(2x)\bigl(kx^{2(n-k)}\bigr)
\qquad\qquad\qquad\qquad
\\[4pt]
=\,&
(2k)x^{2(n-k)+1}
\\[4pt]
=\,&
(2n+1-m)x^m
\end{align*}
Thus in both cases, the $x^m$ term in the expansion of the RHS matches the $x^m$ term of the LHS, so the verification is complete.
