$2n\choose0$ - $2n\choose2$ + $2n\choose4$ - ... + $(-1)^n$$2n\choose2n$
Q: What's the value for any given n $\in$ $N$?
Task is quite easy for odd n and I completed it. You just have to notice that the first number and the last one are opposite etc. I have a problem with even numbers. I noticed it will be $2^n$, but not always positive. Any ideas?
On
Precalculation + OEIS http://codepad.org/tjwaQvuT https://oeis.org/search?q=-4%2C16%2C-64%2C256%2C-1024%2C4096%2C-16384%2C65536%2C-262144%2C1048576%2C
Answer maybe is (-4)^n
The given expression is the real part of $(1+i)^{2n}$. \begin{align*} (1+i)^{2n} &= \left(\sqrt{2}(\cos \pi/4 + i \sin\pi/4)\right)^{2n} \\ &= 2^n (\cos n\pi/2 + i\sin n\pi/2) \end{align*} Thus when $n$ is odd, the sum is zero and when $n = 2k$, sum equals $(-1)^k 2^{2k}$