Binomial Theorem Identity

Question

Using the binomial theorem, derive the following identity:

$\dbinom{n}{0} - \dbinom{n}{1} \frac{1}{2} + \dbinom{n}{2} \frac{1}{4} - ... \pm \dbinom{n}{n} \frac{1}{2^n} = \frac{1}{2^n}$

It is easy to derive $\dbinom{n}{0} + \dbinom{n}{1} + \dbinom{n}{2} + ... + \dbinom{n}{n} = {2^n}$ using pascal's triangle. However, I am confused how to do "one over" this identity.

Answer 1

In general,

$$(x + a)^n = {n \choose 0}x^n + {n \choose 1} x^{n-1} a + \dots + {n \choose n-1} x a^{n-1} + {n \choose n} a^n$$

Your knowledge that $\displaystyle \sum_{k = 0}^{n} {n \choose k} = 2^k$ is a direct consequence of this, namely when $x = a = 1$, the sum is equal to $(1 + 1)^n = 2^n$

Here, we have $x = 1$ and $a = -\frac{1}{2}$.