Problem: Find a closed-form expression for $ \binom{n}{1}+3\binom{n}{3}+5\binom{n}{5}+7\binom{n}{7}+\cdots $ Solution: The above is just $\sum_{k\geq_{0}} (2k+1) {n \choose 2k+1}$ and we see that this counts the number of odd groups of people from among $n$ such that one of them is the president (we are summing over all possible odd numbers that are less than $n$ and when $k > \frac{n-1}{2}$ the binomial term is $0$). Another way to count this is to appoint the president at the start and choose any even number of people from among the $n-1$ to also be in the group. There are $n$ ways to choose the president. Then, the number of even subsets from the remaining $n-1$ people is $2^{n-2}$ because we make a binary choice whether or not to include each person and then the last person either must be in the group or must not be by parity. Hence, $ \sum_{k\geq_{0}} (2k+1) {n \choose 2k+1} = n{2}^{n-2} $