Can you prove this equality? Binomial coefficients and probability frequency

Question

Here's the equality : $$\frac{\sum_{k=0}^n \binom{n}{k}(k/n)}{2^n} = 1/2$$

I've tried for 1, 2, 3 and found the equality was right. I don't know how to prove it (by induction, maybe, but the formula looks too complicated for induction to me) and I'd also like to know (if there is any) the name of this equality. I found it when looking at the probability of a coin (2 sides) and I took the mean of all the frequencies weighted by their amount of occurence. Could have been a dice (6 sides) so I feel like this formula could exist for 1/6 = .... That's why I am looking for the generic name of it !

Thank you in advance for your help in this problem

Answer 1

Your formula, written under the form

$$\sum_{k=0}^n k \frac{\binom{n}{k}}{2^n} = \dfrac{n}{2}$$

plainly expresses that the mean of the binomial probability distribution Bin($n,p=\frac12$) is $np=\dfrac{n}{2}.$

Answer 2

Just differentiate the identity $$ \sum_{k=0}^n \binom{n}{k} x^k=(1+x)^n $$ and get $$ \sum_{k=0}^n k \binom{n}{k} x^k=n (1+x)^{n-1}. $$ Then put $x=1$: $$ \sum_{k=0}^n k \binom{n}{k}=n 2^{n-1}, $$ or $$ \frac{1}{2^n}\sum_{k=0}^n \binom{n}{k} \frac{k}{n} =\frac{1}{2}, $$ as required.

Answer 3

Note that

$$\binom n k \frac k n = \frac{n!}{k!(n-k)!} \frac k n = \frac{(n-1)!}{(k-1)!(n-k)!} = \binom{n-1}{k-1}$$

It is known that

$$\sum_{k=0}^n \binom n k = 2^n$$

Hence, naturally,

$$S_n := \sum_{k=0}^n \binom n k \frac k n =\sum_{k=1}^n \binom n k \frac k n = \sum_{k=1}^{n} \binom{n-1}{k-1}$$

Reindex according to $m=k-1$. Then

$$S_n = \sum_{m=0}^{n-1} \binom{n-1}{m} = 2^{n-1}$$

This is essentially identical to your identity after division by $2^n$.