Subtracting even and odd binomial coefficients? What number do I get if I subtract the binomial coefficients $n\choose k$ with an even $k$ from those with an odd $k$, where $n$ is fixed? Am I supposed to subtract a binomial coefficient with an even $k$ from one with an odd $k$ and say what the number you would get or how exactly does this work? 
 A: You get $0$.  The quickest way to see that is via the binomial theorem:
$$
\sum_{k=0}^n \binom n k (-1)^k 1^{n-k} = ((-1)+1)^n = 0.
$$
There is also a combinatorial argument: Choose a distinguished element from a set of $n$ elements.  Every subset of the set of $n$ elements either contains the distinguished element or does not.  Pair them off: each set containing the distinguished element is paired with the set you get from it by deleting the distinguished element.  One of those two sets has an odd number of elements and the other an even number.  (In some cases the one with the distinguished element is the one with an even number and in some cases it's the one with an odd number.)  This shows that there are exactly as many subsets of even size as of odd size. Therefore when one is subtracted from the other, you get $0$.
Here's a way to do it by thinking about Pascal's triangle.  You add a row to its one-place horizontal shift to get the next row, thus:
$$
\begin{array}{cccccccccccccccccccccccccc}
& & & 1 & & 5 & & 10 & & 10 & & 5 & & 1 \\
+ & 1 & & 5 & & 10 & & 10 & & 5 & & 1 \\[10pt]
\hline
& 1 & & 6 & & 15 & & 20 & & 15 & & 6 & & 1
\end{array}
$$
Now suppose the signs alternate:
$$
\begin{array}{cccccccccccccccccccccccccc}
& & - & 1 & + & 5 & - & 10 & + & 10 & - & 5 & + & 1 & = & 0\text{ ?} \\
+ & 1 & - & 5 & + & 10 & - & 10 & + & 5 & - & 1 & & & = & 0 \text{ ?} \\[10pt]
\hline
+ & 1 & - & 6 & + & 15 & - & 20 & + & 15 & - & 6 & + & 1 & = & \text{?}
\end{array}
$$
If making the signs alternate in row $5$ makes the sum equal to $0$, then the sum in row $6$ must be $0$ since it's just the sum of two $0$s.  If it works in one row, then it works in the next row, so essentially this is a proof by mathematical induction (starting with row $1$, since it doesn't work in row $0$).
A: $$(1+x)^n=\binom{n}{0}+\binom{n}{1}x+\binom{n}{2}x^2+\cdots+\binom{n}{n}x^n$$
Put $x=-1$.
