<s>[INST] 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
 [/INST] 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 $$C_E=2n+1$$ 
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