sum of coefficient of all even power of $x$ 
The sum of all Coefficient of even power of $x$ in
$(1-x+x^2-x^3+\cdots +x^{2n})(1+x+x^2+x^3+\cdots +x^{2n}),n\in \mathbb{N}$

what i try
for $n=1,$ we have
$(1-x+x^2)(1+x+x^2)=(1+1)-(1)+(1+1)=3$
for $n=2,$ we have
$(1-x+x^2-x^3+x^4)(1+x+x^2+x^3+x^4)$
$=(1+1+1)-(1+1)+(1+1+1)-(1+1)+(1+1+1)=5$
so in this way , get  sum of coefficient in original expression is $2n+1$
but How do i solve it without substituting value of $n$, Help me

$$f(x)=(1-x+x^2-x^3+\cdots +x^{2n})(1+x+x^2+x^3+\cdots +x^{2n})$$
$$f(x)=\Bigr[(1+x^2+\cdots +x^{2n})-(x+x^3+\cdots +x^{2n-1})\Bigr]\times \Bigr[(1+x^2+\cdots +x^{2n})+(x+x^3+\cdots +x^{2n-1})\Bigr]$$
$$f(x)=(1+x^2+\cdots +x^{2n})^2-(x+x^3+\cdots +x^{2n-1})^2$$
$$f(x)=\biggr(\frac{1-x^{2n}}{1-x^2}\biggr)^2-\biggr(\frac{x-x^{2n-1}}{1-x^2}\biggr)^2$$
$$f(x)=\frac{1+x^{4n}-2x^{2n}-x^2-x^{4n-2}+2x^{2n}}{(1-x^2)^2}$$
$$f(x)=\frac{1-x^2+x^{4n}-x^{4n-2}}{(1-x^2)^2}$$
$$f(x)=\frac{(1-x^2)(1-x^{4n-2})}{(1-x^2)^2}$$
$$f(x)=\frac{1-x^{4n-2}}{1-x^2}$$
