Be $n\in\mathbb{N}_0$ .
This is a simple formula but not obviously :
$$\sum\limits_{k=0}^{\lfloor{n/2}\rfloor}(-1)^k\binom{n-k}{k}=\frac{2}{3}\left(\cos\frac{\pi(n-1)}{3}+\cos\frac{\pi n}{3}\right)$$
Can be found a short proof for that ?
$\sum\limits_{k=0}^{\lfloor{n/2}\rfloor}(-1)^k\binom{n-k}{k}=\frac{2}{3}\left(\cos\frac{\pi(n-1)}{3}+\cos\frac{\pi n}{3}\right)$
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real-analysis
discrete-mathematics
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0The current version of the question does not make sense. Sum of what? Is the first $=$ not meant to be there? – 2017-01-27
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0@Jan: Thanks a lot for the hint, was only a mistake. – 2017-01-27
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0@FelixMarin : Yes, thanks for the hint, I haven't known this link. :-) I have marked it as double now. – 2017-01-27
1 Answers
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Chebyshev polynomials of the second kind have the following representation: $$ U_n(x)=\sum_{r\geq 0}\binom{n-r}{r}(-1)^r (2x)^{n-2r} \tag{1}$$ hence the wanted sum is just $U_n\left(\frac{1}{2}\right)$, and since $\frac{1}{2}=\cos\frac{\pi}{3}$, $$ U_n\left(\frac{1}{2}\right) = \frac{\sin((n+1)\pi/3)}{\sin(\pi/3)}\tag{2} $$ and the claim is a straightforward consequence of $(2)$.
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1As often, I am amazed at the variety of knowledge you have. I have been working on the $U_n(x)$ but I had not paid attention to their expression. – 2017-01-27
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1Thanks for your nice explanation. :-) ( It's different to the link above which I haven't known. ) – 2017-01-27