How do we get to the following results:

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


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

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

Thank you very much.


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

Can you do the other one ? ;)

  • 1
    $\begingroup$ So we would have $\sum_{i=0}^n 2^{-3i} {n \choose i} = \sum_{i=0}^n {n \choose i} \left(\frac{1}{8}\right)^i 1^{n-i} = \left(1+ \frac{1}{8}\right)^n = \left(\frac{9}{8}\right)^n$. This is very nice! Thank you for your help $\endgroup$ – user136457 Nov 7 '14 at 19:30
  • 1
    $\begingroup$ @user136457 That same reasoning is used to show that $\sum_{k=0}^n\binom{n}{k}=2^n$ by the way, can you see why ? ;) $\endgroup$ – Hippalectryon Nov 7 '14 at 19:31
  • $\begingroup$ So I suppose because $\sum_{k=0}^n {n \choose k}=\sum_{k=0}^n {n \choose k} 1^k 1^{n-k} = (1+1)^n=2^n$. This is a very nice trick! Thank you for showing it to me. $\endgroup$ – user136457 Nov 7 '14 at 19:39

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.


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