I am given the question:
What is the value of the following? $$\sum\limits_{k=0}^5\binom{5}{k}4^{5-k}(-2)^k$$
I missed class when the teacher went over a similar problem, but I know we have to use the binomial theorem. I'm not sure which value she means we have to get...
It is $\sum_{k=0}^5 \binom {5}{k}4^{5-k} (-2)^k = (4+(-2))^5 = 2^5 $. See also here https://en.wikipedia.org/wiki/Binomial_theorem.