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
 A: These are two standard geometric series,
one with ratio $-x$,
the other with ratio $x$.
Their product is
$(1-(-x)^{2n+1})(1-x^{2n+1})/((1+x)(1-x))\\
=(1+x^{2n+1})(1-x^{2n+1})/(1-x^2)\\
=(1-x^{4n+2})/(1-x^2)
$
At $x=1$, applying Hoppy,
we get
$(4n+2)x^{4n+1}/(2x)
=(2n+1)x^{4n}
\to 2n+1
$.
A: $$f(x)=(1-x+x^2-x^3+\cdots +x^{2n})(1+x+x^2+x^3+\cdots +x^{2n})$$
$$f(x)=\Bigr[(1+x^2+\cdots +x^{2n})-(x+x^3+\cdots +x^{2n-1})\Bigr]\times \Bigr[(1+x^2+\cdots +x^{2n})+(x+x^3+\cdots +x^{2n-1})\Bigr]$$
$$f(x)=(1+x^2+\cdots +x^{2n})^2-(x+x^3+\cdots +x^{2n-1})^2$$
$$f(x)=\biggr(\frac{1-x^{2n}}{1-x^2}\biggr)^2-\biggr(\frac{x-x^{2n-1}}{1-x^2}\biggr)^2$$
$$f(x)=\frac{1+x^{4n}-2x^{2n}-x^2-x^{4n-2}+2x^{2n}}{(1-x^2)^2}$$
$$f(x)=\frac{1-x^2+x^{4n}-x^{4n-2}}{(1-x^2)^2}$$
$$f(x)=\frac{(1-x^2)(1-x^{4n-2})}{(1-x^2)^2}$$
$$f(x)=\frac{1-x^{4n-2}}{1-x^2}$$
A: Let's call sum of odd coefficients $C_O$ and sum of even coefficients $C_E$. You can write your polynomial as $$P(x)=O(x)+E(x)$$, where $O(x)$ is the sum of odd powered terms and $E(x)$ is the sum of even powered terms. Note that if you plug in $x=1$, you get $$C_O=O(x=1)$$ and $$C_E=E(x=1)$$ Similarly, if you plug in $x=-1$, the odd terms will acquire a minus sign, while the even ones will be unchanged. So $$O(x=-1)=-C_O$$ and $$E(x=-1)=C_E$$
Then $$C_E=\frac12 (E(1)+E(-1))=\frac12(E(1)+E(-1))+\frac12(O(1)+O(-1))=\frac12(P(1)+P(-1))$$ 
If you plug in $x=1$ in your original form, the second parenthesis will evaluate to $2n+1$, and the first parenthesis will be $1$. If you put in $x=-1$, you have the first parenthesis $2n+1$ and the second one $1$. Then $$C_E=2n+1$$ 
