<s>[INST] Find the value of $a_{0}^{2}-a_{1}^{2}+a_{2}^{2}- \dots+a_{2 n}^{2}$ Let n be a positive integer and $$\left(1+x+x^{2}\right)^{n}=a_{0}+a_{1} x+\cdots+a_{2 n} x^{2 n}$$
then the value of $a_{0}^{2}-a_{1}^{2}+a_{2}^{2} - \dots+a_{2 n}^{2}$ is
My approach:-
Replacing $x$ by $(-1 / x),$ we get
$$
\begin{array}{r}
\left(1-\frac{1}{x}+\frac{1}{x^{2}}\right)^{n}=a_{0}-\frac{a_{1}}{x}+\frac{a_{2}}{x^{2}}+\cdots-a_{2 n-1} \cdot \frac{1}{x^{2 n-1}}+\frac{a_{2 n}}{x^{2 n}} \\
\text { or, }\left(1-x+x^{2}\right)^{n}=a_{0} x^{2 n}-a_{1} x^{2 n-1}+a_{2} x^{2 n-2}+\cdots+a_{2 n}..... \tag{1}
\end{array}
$$
And given $\left(1+x+x^{2}\right)^{n}=a_{0}+a_{1} x+\cdots+a_{2 n} x^{2 n} \ldots \ldots \ldots \ldots \ldots . \tag{2}.$
Multiplying corresponding sides of (1) and $(2),$ we have
$$
\left(1+x^{2}+x^{4}\right)^{n}=\left(a_{0}+a_{1} x+a_{2} x^{2}+\cdots+a_{2 n} x^{2 n}\right) \times\left(a_{0} x^{2 n}-a_{1} x^{2 n-1}+a_{2} x^{2 n-2}+\cdots+\right.
$$
$\left.a_{2 n}\right) \ldots \ldots...\tag{3}$
$$
\left(1+x^{2}+x^{4}\right)^{n}=\left(a_{0}+a_{1} x^{2}+a_{2} x^{4}+\cdots+a_{n} x^{2n}+\cdots+a_{2 n} x^{4 n}\right) \ldots \ldots\tag{4}
$$
Equating coefficient of $x^{2 n}$ on both sides of (3) and (4)
$$
a_{0}^{2}-a_{1}^{2}+a_{2}^{2} -\cdots +a_{2 n}^{2}=a_{n}
$$
But this method seems  very tedious to me.
Any other approach would be greatly appreciated
 [/INST] The following is essentially the idea in your proof which is conceptually simple. One has by the given, $$(1+x+x^2)^n=a_0+a_1x+\cdots+a_{2n}x^{2n}.\quad (1)$$
Replacing $x$ by $1/x$ and multiplying by $x^{2n}$ in (1), one sees that $$a_k=a_{2n-k}, 0\leq k\leq 2n.\quad (1)$$
Replacing $x$ by $-x$ in (1), one has $$(1-x+x^2)^n=a_0-a_1x+\cdots+a_{2n}x^{2n}.\quad (2)$$
Replacing $x$ by $x^2$ in (1), one has $$(1+x^2+x^4)^n=a_0+a_1x^2+\cdots a_n x^{2n}+\cdots+a_{2n}x^{4n}.\quad (3)$$
Since $1+x^2+x^4=(1+x+x^2)(1-x+x^2)$, multiplying (2) and (1) and comparing coefficients of $x^{2n}$ with (3), one has $$a_0a_{2n}-a_1a_{2n-1}+a_2a_{2n-2}+\cdots+a_{2n}a_0=a_n,$$ which after applying (1) yields $$a_0^2-a_1^2+a_2^2-\cdots+a_{2n}^2=a_n,$$ as required.
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