Question
if $a_k$ is multinomial coefficient of $x^k$ in polynomial $(1+x+x^2)^n$,where $0\le k\le 2n$,prove:
using this equality $(1+x+x^2)(1-x+x^2)=1+x^2+x^4$,show that $$a^2_o-a^2_1+a^2_2-...+(-1)^{n-1}a^2_{n-1}=\frac {1}{2}(a_n+(-1)^{n+1}a^2_n)$$
thing i have done so far:
$a_k=\frac {n!}{a!b!c!}$,where $a+b+c=n$ and $b+2c=k$
$b_k$ is multinomial coefficient $(1-x+x^2)$,$b_k=\frac {(-1)^bn!}{a!b!c!}$,where $a+b+c=n$ and $b+2c=k$
when $k$ is even , $b$ is even too,so $b_k=a_k$ (1)
when $k$ is odd , $b$ is odd too,so $b_k=-a_k$ (2)
$a_k$ is multinomial coefficient of $x^k$ in $(1+x^2+x^4)^n$
in this equality $(1+x+x^2)^n(1-x+x^2)^n=(1+x^2+x^4)^n$ , multinomial coefficient $x^{2n}$ is $a_n$ on right side and on left side is equal to:$$a_0b_{2n}+a_1b_{2n-1}+a_2b_{2n-2}+...+a_{2n}b_0$$ (3)
i proved (in this question): $$a_{n+k}=a_{n-k}$$ (4)
so using (1),(2),and (4) i can say:
$b_{2n}=a_{2n}=a_0$ , $b_{2n-1}=-a_{2n-1}=a_1$ , $b_{2n-2}=a_{2n-2}=a_2$, ...
then i can rewrite (3):
$a^2_o-a^2_1+a^2_2-...+a^2_{2n}$
and i stuck.
The equation at hand can be rewritten as
$$2(a_0^2 - a_1^2 + a_2^2 + \cdots + (-1)^{n-1} a_{n-1}^2) + (-1)^n a_n^2 \stackrel{?}{=} a_n$$
Since $a_k = a_{n-2k}$ for $k : 0 \le k \le 2n$, the LHS can be rewritten as
$$a_0 a_{2n} - a_1 a_{2n-1} + a_2 a_{2n-2} + \cdots + a_{2n-2} a_2 - a_{2n-1} a_1 + a_{2n} a_0$$ This is the coefficient in front of $x^{2n}$ in the expansion
$$(a_0 - a_1 x + a_2 x^2 + \cdots )(a_0 + a_1 x + a_2 x^2 + \cdots ) = (1 - x + x^2)^n ( 1 + x + x^2 )^n \\ = ( 1 + x^2 + x^4 )^n = a_0 + a_1 x^2 + a_2 x^4 + \cdots + a_n x^{2n} + \cdots $$ and hence equal to $a_n$ in RHS.