If $(1+x+x^2)^n=a_0+a_1x+a_2x^2+\dots+a_{2n}x^{2n}$ then prove that $a_0^2-a_1^2+a_2^2-a_3^2+\dots+(-1)^{n-1}a_{n-1}^2=\frac12a_n(1-(-1)^na_n)$

Question

If $(1+x+x^2)^n=a_0+a_1x+a_2x^2+\dots+a_{2n}x^{2n}$ then prove that $a_0^2-a_1^2+a_2^2-a_3^2+\dots+(-1)^{n-1}a_{n-1}^2=\frac12a_n(1-(-1)^na_n)$

My Attempt:

Replacing $x$ by $-x$ in the given expression, $$(1-x+x^2)^n=a_0-a_1x+a_2x^2-\dots+a_{2n}x^{2n}$$

Therefore, $$(1+x+x^2)^n(1-x+x^2)^n=(a_0+a_1x+a_2x^2+\dots+a_{2n}x^{2n})(a_0-a_1x+a_2x^2-\dots+a_{2n}x^{2n})$$

LHS $=((1+x^2)^2-x^2)^n=(1+x^2+x^4)^n$

And if from the right side, I try to find the coefficient of $x^{2n}$, I get $a_0a_{2n}-a_1a_{2n-1}+\dots$

It seems like I need to do $a_0=a_{2n}, a_1=a_{2n-1}, \dots$

$a_0=^nC_0$ and $a_{2n}=^nC_n$, so, they are equal.

I am confused with $a_{n-1}$ and $a_{n+1}$.

$a_{n+1}=^nC_{\frac{n+1}2}$. But for this, $n$ should be odd, right? What if $n$ is even?

For $a_{n-1}$, would we obtain coefficients both from $x$ and $x^2?$

Answer 1

Your approach is absolutely the way to go. Just a tiny step is missing.

So, you let $f(x)=(1+x+x^2)^n=\sum_{k=0}^{2n}a_kx^k$ and compute the coefficient of $x^{2n}$ in $$f(x)f(-x)=f(x^2)$$ in two ways: obviously $[x^{2n}]f(x^2)=a_n$; on the other hand, $$[x^{2n}]f(x)f(-x)=\sum_{k=0}^{2n}(-1)^k a_k a_{2n-k}=\sum_{k=0}^{2n}(-1)^k a_k^2=(-1)^na_n^2+2\sum_{k=0}^{n-1}(-1)^ka_k^2;$$ the last two equalities follow from $a_k=a_{2n-k}$ which is a consequence of $f(x)=x^{2n}f(1/x)$.