Show that $$\sum_{i=0}^{n}\binom{k}{i}=2^{k-1}$$ for every positive odd integer in the form of $k=2n+1$ ($n$ being a positive integer).
I tried the usual induction demonstration.
The case for $k=1$ is rather trivial. I'm currently stuck at the second part of the proof. I know that $\binom{k}{0}=1$ for every $k$ and $$ \sum_{i=0}^{n} \binom{n}{i} = 2^n$$ but despite that I'm having trouble arriving at the desired result.
This follows from the symmetry in the rows of Pascal's Triangle.
Namely, the last $n+1$ numbers in row $k=2n+1$ of Pascal's Triangle are just the first $n+1$ numbers in reverse order, and so the sum of the first $n+1$ numbers ($\binom{k}{0}$ to $\binom{k}{n}$) is half of the sum of all the numbers in row $k$, which is equal to $\frac{1}{2} \cdot 2^k=2^{k-1}$.