Proof $\dbinom{n}{0} - \frac{1}{2}\dbinom{n}{1} + \cdots + (-1)^n\frac{1}{2^{n-1}}\dbinom{n}{n-1}$

Question

For all $n \ge 1$, $$\binom{n}{0} - \frac{1}{2}\binom{n}{1} + \frac{1}{2^2}\binom{n}{2} - \frac{1}{2^3}\binom{n}{3} + \cdots + (-1)^n\frac{1}{2^{n-1}}\binom{n}{n-1} = 0,$$ if $n$ is even                                                                                   $$\frac{1}{2^{n-1}}, \text{ if } n \text{ is odd}$$

I need to use the binomial theorem in this question.

Answer 1

The given sum is $$ \sum_{k=0}^{n-1}\binom{n}{k}\left(-{1\over2}\right)^k=\sum_{k=0}^{n}\binom{n}{k}\left(-{1\over2}\right)^k-\left(-{1\over2}\right)^n=\left(1-{1\over2}\right)^n-\left(-{1\over2}\right)^n\\=\left({1\over2}\right)^n-\left(-{1\over2}\right)^n $$

Answer 2

Try to use the binomial formula on $(1+(-\frac{1}{2}))^n$
This will give you to the proof of the statement.