Calculating the sum of coefficients for binomial expression The question is:
Calculate the sum of the coefficients of $(a-b)^{250}$.
My reasoning was that we can take the example of $(a-b)^2$, which would have the coefficients of $1$, $-2$, and $1$, according to Pascal's triangle. This would add up to $0$. 
Then, to calculate the sum of coefficients for the original problem, $((a-b)^2)^{125}$ would equate to $(a-b)^{250}$, and since the sum of the coefficients of $(a-b)^2$ is $0$, then $(0)^{125}$ would be $0$, so the sum of the coefficients of $(a-b)^{250}$ would be 0.
Is my methodology correct? If not, is there another methodology that is better?
 A: HINT: If $f(a,b)$ is a polynomial in $a$ and $b$ of the form
$$f(a,b) = \sum_{k=0}^n c_k a^kb^{n-k}$$
the sum of coefficients is given by $f(1,1) = \displaystyle\sum_{k=0}^n c_k$.
A: Hint:  Try a few more small cases and see what happens (keep using Pascal's triangle... note that the signs of each term alternate beginning with a positive sign).
$1, -2, 1 \rightarrow 0$  This is $n=2$.
$1, -3, 3, -1 \rightarrow 0$  This is $n=3$.
$1, -4, 6, -4, 1 \rightarrow 0$  This is $n=4$.
$1, -5, 10, -10, 5, -1 \rightarrow 0$  This is $n=5$.
$1, -6, 15, -20, 15, -6, 1 \rightarrow 0$  This is $n=6$.
The case when $n$ is odd is fairly easy because $\binom{n}{i}=\binom{n}{n-i}$ and exactly one of those will be positive and exactly one will be negative (so they cancel).  What happens when $n$ is even is a little more subtle.
You would be done if you could show why $\sum \binom{n}{2i} = \sum \binom{n}{2i+1}$ when $n$ is even.  There is a nice combinatorial way to do this, but I'll let you look for it.
A: Just write down the general formula for the binomial expansion of $(a-b)^n$. Then consider what happens when $a = b$. You will see that the sum of the binomial coefficients must be equal to zero. This result is true for every $n > 0$.  
