The identity
[tex]\displaystyle \sum k \binom{n}{k} = n 2^{n-1} [/tex]
is pretty standard, and one can prove it algebraically by cancelling the k in the sum with the binomial coefficient and then using the binomial theorem summation or a combinatorial proof that goes like this.
Choose a k element committee with from n members with one of the comittee member being made the chairperson.
A third proof using probability goes like this. First divide throughout by [tex]2^n[/tex]. Next consider tossing n fair coins with probability [tex]p=\frac{1}{2}[/tex] of turning up heads or tails. The expected number of heads is [tex]\frac{n}{2}[/tex]. On the other hand, the probability of having exactly k heads (and hence n-k tails) is [tex]\binom{n}{k} (p)^k (1-p)^{n-k}[/tex].
Thus the expected value is
[tex] \displaystyle \sum k \binom{n}{k} (\frac{1}{2})^{n}= \frac{n}{2}. [/tex]
From a probabilistic point of view, it could very well be the fact that one proves the expected value of is np by using the identity so this third proof isn’t really a proof but a nice application.