Binomial theorem alternating sum I come, once again looking for a tutorial.  It seems like my discrete math class is lacking in explanations for more complex (to me) problems.  My question is:  
Compute the value of the following alternating sum and show all your work.  
$$\binom{30}{3}4-\binom{30}{4}16+\binom{30}{5}64\mp \cdots-\binom{30}{30}4^{28}$$  
Leave expressions of the form $m^n$ in your answer without attempting to evaluate them.  Solutions that rely solely on calculator computation of the terms of the sum are unacceptable.  
End Question  
Please do not answer the exact question, I would like to find the solution on my own.  If someone could only show how to work this problem, I would be grateful.  Thank you.
 A: \begin{align}
& \binom{30}{3}4-\binom{30}{4}16+\binom{30}{5}64\mp \cdots-\binom{30}{30}4^{28} \\[10pt]
 = {} & \frac {-1} {16} \left( \binom{30}{3}(-4)^3+\binom{30}{4}(-4)^4+\binom{30}{5} (-4)^5 + \cdots+\binom{30}{30}(-4)^{30} \right)\\[10pt]
 = {} & \frac {-1} {16} \left( \underbrace{\binom{30}{0}(-4)^0+\binom{30} 1 (-4)^1 + \binom{30} 2 (-4)^2 + \cdots+\binom{30}{30}(-4)^{30}} - \, \text{(three terms)} \right)
\end{align}
The binomial theorem handles the part over the $\underbrace{\text{underbrace}}$.
A: Think of the binomial theorem...
$$(x+y)^n=\sum_{k=0}^n\binom{n}{k}x^ky^{n-k}$$
clearly there is an alternating sum, so we can let $y=-1$.  That yields
$$(x-1)^n=\sum_{k=0}^n\binom{n}{k}(-1)^{n-k}x^k$$
Now, consider what $n$ is and consider what $x$ is... you may have to subtract terms off the summation, and consider a scalar $b$...
$$b(x-1)^{30}=\sum_{k=0}^{30}\binom{30}{k}(-1)^{30-k}x^k$$
Note that also when $k$ is odd, we have positive terms and when $k$ is even we have negative terms...that should set off bells of what $b$ should be...
