Is there a quick, fancy, way of proving sums such as this?
Prove that: $$\sum_{x=0}^{n} (-1)^x {n \choose x} = 0$$
A recent homework assignment I turned in had a couple problems similar to the above. For the most part, I used a proof by induction to solve them. They take forever to write up that way and I was wondering if there were any manipulations possible to speed these types of problems up.
Any suggestions would be greatly appreciated!
$$(1-1)^n=\sum_{x=0}^n {n\choose x}(-1)^x=0$$
Conisder the following expansion $$(x-a)^n=\sum_{i=0}^n{n \choose i}x^{n-i}(-a)^i$$. . In your question $x =1$ and $a=1$.
Many of theses can be obtained as expansions of $(L+K)^n$. Usually, L=1. Here K=-1, Sometimes you need K=1, or $\omega$ or $\omega^2$
Amrutam's proof is absolutely correct and very simple, but I won't give it away if he won't. Instead I'll give you a different proof. You can actually use this proof to generalize alternating sums on a row in Pascal's Triangle. Anyway, by Pascal's Identity we get that
$\sum\limits_{i=0}^{n} (-1)^{i}{n \choose i}=\sum\limits_{i=0}^{n} (-1)^{i} \left({n-1 \choose i}+{n-1 \choose i-1}\right)$
Notice this is a telescoping sum and...
