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!
Calculating a binomial sum
3
$\begingroup$
discrete-mathematics
binomial-coefficients
generating-functions
2 Answers
11
Hints: $(-1)^k=(-1)^{10-k}$ and binomial theorem.
-
2Alternatively, combine the powers into $(-1/2)^k$ and put a $1^{n-k}$ alongside it. – 2012-03-12
-
0Sorry, I didn't get it. Can you be more specific? – 2012-03-12
-
0@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
-
0@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
-
0@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
3
Another Hint
$$\sum_{k=0}^{n} \binom{n}{k} x^k y^{n-k} = (x+y)^n$$
-
0Just realized that anon mentioned Binomial Theorem (and here is my silly hint) – 2012-03-12