This question is similar to some questions I've already seen but it's still hard for me to grasp how to do this. I have noted that
$(1+1)^{2n} = \sum_{k=0}^{2n} {2n \choose k}$
and
$(1-1)^{2n} = \sum_{k=0}^{2n} {2n \choose k}(-1)^k$
where the first argument gives the sum of all even and odd indexes while the second gives the difference between the sum of all even indexes and odd indexes. I can see that for the purposes of this problem we are only concerned with the even indexes and so I thought about
$(1+4)^{2n} = \sum_{k=0}^{2n}{2n \choose k}4^k$
having all even and odd indexes but translating the problem above with $x=1$ in $(1+x)^{2n}$, I can't use
$(1-4)^2n = \sum_{k=0}^{2n}{2n \choose k}(-4)^k$ and add this to $(1+4)^{2n}$ to find all even indexes since this expression doesn't divide up the even and odd indexes as $(1-1)^{2n}$, (-4 is even). I also noted that the problem can be expressed as
$\sum_{k=0}^n{2n \choose 2k}2^{2k}$
which I thought could get me somewhere as now the bottom term of the coefficient perfectly aligns with the exponent, but I'm still stuck on how to do this.
Note: This question was assigned as review for an exam not homework. Not expecting a full solution, hints would be great.
I don't see a problem with the approach you just explained. $$\begin{align} (1+2)^{2n}+(1-2)^{2n} &=\sum_{k=0}^{2n}\binom{2n}{k}2^k+\sum_{k=0}^{2n}\binom{2n}{k}(-2)^k\\ &=\sum_{k=0}^{2n}\binom{2n}{k}(2^k+(-2)^k)\\ &=2\sum_{j=0}^n\binom{2n}{2j}2^{2j}\quad\left(j=\frac12k\right)\\ &=2\sum_{j=0}^n\binom{2n}{2j}4^j\\ \end{align}$$ $$\therefore\sum_{k=0}^n\binom{2n}{2k}4^k=\frac12\left(3^{2n}+(-1)^{2n}\right)$$