Solve easy sums with Binomial Coefficient

Question

How do we get to the following results:

$$\sum_{i=0}^n 2^{-i} {n \choose i} = \left(\frac{3}{2}\right)^n$$

and

$$\sum_{i=0}^n 2^{-3i} {n \choose i} = \left(\frac{9}{8}\right)^n.$$

I guess I could prove it by induction. But is there an easy way to derive it?

Thank you very much.

Answer 1

$\displaystyle\sum_{i=0}^n 2^{-i} {n \choose i}=\sum_{i=0}^n {n \choose i}\left(\frac{1}2\right)^i\cdot1^{n-i}= \left(1+\frac{1}2\right)^n$

Can you do the other one ? ;)

Answer 2

For the first one consider the binomial expansion of $(1+\frac{1}{2})^n$ and see how close that is the left hand side while adding the values will give the right hand side.

For the second, consider putting in $\frac{1}{8}$ and note what fraction is 2 to the negative 3.