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I came across this excercise in an old exam (in discrete math), and I don't know how to approach it: $$\sum_{k=0}^{10}\left(\frac{1}{2}\right)^k\left(-1\right)^k\binom{10}{k}$$ I know the answer is $2^{-10}$, but I don't know why. When I was going through my text book I saw something similar regarding Catalan numbers generating functions. Thanks!

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Hints: $(-1)^k=(-1)^{10-k}$ and binomial theorem.

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    Alternatively, combine the powers into $(-1/2)^k$ and put a $1^{n-k}$ alongside it.2012-03-12
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    Sorry, I didn't get it. Can you be more specific?2012-03-12
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    @yotamoo: Did you look at the link? Do you see why $(-1)^k=(-1)^{n-k}$? Do see why substituting the former with the latter in your expression gives you **exactly the binomial theorem for** $(1/2-1)^{10}$ **?**2012-03-12
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    @anon: Actually I'd say it is $(-\frac12+1)^{10}$ by the binomial theorem; this avoids shuffling around minus signs. Not a big deal, admitted.2012-03-12
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    @Marc: $(1/2-1)^{10}$ is the form in my answer, $(-1/2+1)^{10}$ is the form in my first comment.2012-03-12
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Another Hint

$$\sum_{k=0}^{n} \binom{n}{k} x^k y^{n-k} = (x+y)^n$$

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    Just realized that anon mentioned Binomial Theorem (and here is my silly hint)2012-03-12