Proof : Sum of k(n choose k)

This article presents the proof of: the sum of k times n choose k = n times 2 to the power of (n minus 1).


One of the famous formulas using binomial coefficients is:

$$\sum^n_{k=1} k\binom{n}{k} = n \times 2^{n-1}$$

The proof:

  1. We start by using Newton’s binomial formula:

$$\sum^n_{k=1} \binom{n}{k}a^{k}b^{n-k}=(a+b)^n$$

  1. Let $b = 1$, then :

$$\sum^n_{k=1} \binom{n}{k}a^{k} = (a+1)^n$$

  1. Differentiate both members of the equation with respect to $a$ like this:

$$\frac{d}{da}(\sum^n_{k=1} \binom{n}{k}a^{k} = (a+1)^n)$$

  1. Which gives us:

$$\sum^n_{k=1} \binom{n}{k}ka^{k-1} = n(a+1)^{n-1}$$

  1. We can now take $a = 1$ and get:

$$\sum^n_{k=1} k\binom{n}{k} = n \times 2^{n-1}$$

Others identities here: