<s>[INST] Inductive proof of $\sum_{i=0}^{n} 2^{-i} \binom{n}{i} = \left(\frac{3}{2} \right)^n$ I am trying to prove by induction on $n$ the following theorem:
$$\sum_{i=0}^{n} 2^{-i} \binom{n}{i} = \left(\frac{3}{2} \right)^n$$
For my inductive step I have:
$$\sum_{i=0}^{n+1} 2^{-i} \binom{n+1}{i}$$
$$\sum_{i=0}^{n} 2^{-i} \binom{n+1}{i} + \frac{1}{2^{-(n+1)}}$$
Using this identity: $\binom{n+1}{i} = \binom n i + \binom{n}{i-1}$:
$$\sum_{i=0}^{n} 2^{-i} \binom{n}{i} + \sum_{i=0}^{n} 2^{-i} \binom{n}{i-1} + \frac{1}{2^{-(n+1)}}$$
$$\left( \frac{3}{2} \right)^n + \frac 1 2\sum_{k=1}^{n} 2^{-k} \binom{n}{k} + \frac{1}{2^{-(n+1)}}$$
$$\left( \frac{3}{2} \right)^n + \frac 1 2 \left[ \left( \frac{3}{2} \right)^n - 1\right]  + \frac{1}{2^{-(n+1)}}$$
$$\left( \frac{3}{2} \right)^{(n+1)} + \frac 1 2 + \frac{1}{2^{-(n+1)}}$$
Here I am stuck and I cannot find my mistake.
 [/INST] The problem is with the subscript handling for
$$\sum_{i=0}^{n} 2^{-i} \binom{n}{i-1}.\tag{1}$$
Expression (1) is equal to 
$$\sum_{i=1}^{n} 2^{-i} \binom{n}{i-1},$$
which is equal to 
$$\frac{1}{2}\sum_{i=0}^{n} 2^{-(i-1)} \binom{n}{i-1}.\tag{2}$$
Let $j=i-1$. We get that (2) is equal to
$$\frac{1}{2}\sum_{j=0}^{n-1} 2^{-j} \binom{n}{j}.$$
Sum  to $n$ instead, but to keep things unchanged subtract $\frac{1}{2^n}$.
We get 
$$\frac{1}{2}\left(\frac{3}{2}\right)^n -\frac{1}{2^{n+1}}.$$
Now we are finished.  For the last term cancels the $\frac{1}{2^{n+1}}$ term you had at the end. And the $\frac{1}{2}\left(\frac{3}{2}\right)^n$ combines with the $\left(\frac{3}{2}\right)^n$ you already have, and we end up with $(1.5)\left(\frac{3}{2}\right)^n$, that is, $\left(\frac{3}{2}\right)^{n+1}$.
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