<s>[INST] Showing that $ 1 + 2 x + 3 x^2 + 4 x^3 + \cdots + x^{10} = (1 + x + x^2 + x^3 + x^4 + x^5)^2$ I was studying a polynomial and Wolfram|Alpha had the following alternate form:
$$P(x) = 1 + 2 x + 3 x^2 + 4 x^3 + 5 x^4 + 6 x^5 + 5 x^6 + 4 x^7 + 3 x^8 + 2 x^9 + x^{10} = (1 + x + x^2 + x^3 + x^4 + x^5)^2$$
Of course, we can verify this through expansion, but if I were a mathematician without access to CAS, how might I notice that this is the case?
I suppose what I'm asking is how one should "see" that $P$ can be simplified to $(1 + x + x^2 + x^3 + x^4 + x^5)^2$? Is it a multinomial thing (which seems a bit too complicated for someone to "notice"), or is there something simpler about the polynomial that one could use to factor it?
 [/INST] The coefficient of $x^n$ ($0\leq n\leq 10$) in the expansion is the number of solutions to
$$
x_{1}+x_{2}=n;\quad 0\leq x_i\leq5.
$$
We can compute this via inclusion exclusion. Let $U$ be the set of all solutions to the previous question in non-negative integers and $A_{i}$ be the set of solutions in $U$ which $x_i>5$. Then
$$
\begin{align}
|A_{1}^c\cap A_{2}^c|&=|U|-|A_{1}|-|A_{2}|+|A_{1}\cap A_{2}|\\
&=(n+1)-2(n-5)[n\geq6]+0
\end{align}
$$
where $[\cdot]$ is the iverson bracket since $A_{1}\cap A_{2}=\varnothing$.
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