Proving that $\sum\limits_{k=0}^{2n}(-1)^k\binom{4n}{2k}=(-4)^n$

Question

I need sum this numerical serie. $\sum\limits_{k=0}^{2n} (-1)^k \begin{pmatrix}4n\\2k\end{pmatrix}$

I know that the result will be $(-4)^n$ but i don't know how can I get it.

Could you help me with it please?

Answer 1

Basic approach. There's probably an easier way to do this, but the way that comes to mind immediately is to use the binomial theorem on the expression

$$ (1+i)^{4n}+(1-i)^{4n} $$

where $i = \sqrt{-1}$. This will give you twice your summation. Then observe that

$$ (1+i)^{4n} = \left[(1+i)^4\right]^n = (-4)^n $$ $$ (1-i)^{4n} = \left[(1-i)^4\right]^n = (-4)^n $$

and the sum of those two is clearly twice your desired value. The factors of $2$ cancel out, and the formula is thus shown.

Answer 2

$\displaystyle \sum_{0 \le k \le 2n}(-1)^k \binom{4n}{2k} = \sum_{0 \le \frac{1}{2} k \le 2n} i^k\binom{4n}{k} = \sum_{0 \le k \le 4n}\binom{4n}{k}i^k = (1+i)^{4n} = [(1+i)^4]^n = (-1)^n 4^n. $