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
 A: 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.
A: We have
$$
S_{\,n} (x) = \left( {1 + x + x^{\,2} } \right)^{\,n}  = \sum\limits_{k = 0}^{2n} {a_{\,n,\;k} x^{\,k} } 
$$
and
$$
S_{\,n} (x) = x^{\,2n} \left( {1 + x^{\, - 1}  + x^{\, - 2} } \right)^{\,n}  = x^{\,2n} S_{\,n} (1/x)\quad  \Rightarrow \,\quad a_{\,n,\;k}  = a_{\,n,\;2n - k} 
$$
therefore the coefficients are symmetric wrt $k=n$
Then
$$
\eqalign{
  & S_{\,n} (x)S_{\,n} ( - x) = \left( {\left( {1 + x + x^{\,2} } \right)\left( {1 - x + x^{\,2} } \right)} \right)^{\,n}  =   \cr 
  &  = \left( {\left( {1 + x^{\,2} } \right)^{\,2}  - x^{\,2} } \right)^{\,n}  = \left( {1 + x^{\,2}  + x^{\,4} } \right)^{\,n}  = S_{\,n} (x^{\,2} ) \cr} 
$$
which implies
$$
S_{\,n} (x)S_{\,n} ( - x) = \sum\limits_{k = 0}^{2n} {\left( {\sum\limits_{j = 0}^k {\left( { - 1} \right)^{\,k - j} a_{\,n,\;j} \,a_{\,n,\;k - j} } } \right)x^{\,k} }
  = \sum\limits_{k = 0}^{2n} {a_{\,n,\;k} x^{\,2k} }  = S_{\,n} (x^{\,2} )
$$
and thus
$$
\sum\limits_{j = 0}^k {\left( { - 1} \right)^{\,k - j} a_{\,n,\;j} \,a_{\,n,\;k - j} }  = \sum\limits_{j = 0}^k {\left( { - 1} \right)^{\,k - j} a_{\,n,\;j} \,a_{\,n,\;2n - \left( {k - j} \right)} }
   = \left\{ {\matrix{  0 & {k\,{\rm odd}}  \cr    {a_{\,n,\;k/2} } & {k\,{\rm even}}  \cr  } } \right.
$$
and in particular for $k=2n$
$$
\sum\limits_{j = 0}^{2n} {\left( { - 1} \right)^{\,j} a_{\,n,\;j} \,a_{\,n,\;j} }  = a_{\,n,\;n} 
$$
