3

$2n\choose0$ - $2n\choose2$ + $2n\choose4$ - ... + $(-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?

Spideyyyy
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2 Answers2

4

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 $(-1)^k 2^{2k}$

  • How do I know it's the real part of this? – Spideyyyy Oct 27 '16 at 08:21
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    $i^{2k} = (-1)^k$ and hence real, and $i^{2k+1} = (-1)^k i$ and is imaginary. Hence the given sum can be written as $\binom{2n}{0} + (i)^2 \binom{2n}{2} + i^4 \binom{2n}{4} + \cdots$. Adding the other terms we get $(1+i)^{2n}$. Now we see the required sum is the real part of this expression. –  Oct 27 '16 at 08:24
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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

kotomord
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