Sum of coefficients of even powers of x in $(1+x)^5 (1+x^2)^5$?

Question

the exact question is asking the sum of coefficients of even powers of x in the expansion of $(1+x+x^2+x^3)^5$ and in the solution this expression is simplified to $(1+x)^5(1+x^2)^5$. The solution to this question says this expression is equal to $(1+5x+10x^2+10x^3+5x^4+x^5)(1+x^2)^5$, this I understood, then it says that required sum of coefficients is equal to $(1+10+5)2^5$. How did they get this?

Answer 1

The second term $(1+x^2)^5$ will have only even powers.
Now you have to calculate the sum of coefficients,
$\Rightarrow$ $2^5=32$ as ${5 \choose 0}+{5 \choose 1}+{5 \choose 2}+{5 \choose 3}+{5 \choose 4}+{5 \choose 5}=2^5$
And calculate the sum of all even powers in the first term i.e. $(1+x)^5$ as sum of 2 even numbers is an even number.
$\Rightarrow$ $(1+10+5)=16$
So answer is $16*32=512$

Answer 2

The sum of coefficients of even power in $P(x)$ equals $$\frac{P(1)+P(-1)}{2}$$ in your case it's $$\frac{0+2^5\cdot 2^5}{2}=2^9=512$$.