How do I solve a binomial theorem with a factor inserted in

Question

I have a given formula which is very similar to the Binomial Formula

$$\sum_{k=0}^n \frac{k}{n} \binom{n}{k} z^k (1-z)^{n-k} = z$$

I have to prove that the above statement is true. As of right now I could only show that if $\frac{k}{n}$ were not there the sum would get the value 1 which I can only use to prove that $\frac{k}{n} \sim z$. Anyone know of any way to help me with this. I have also tried solving with induction but I get stuck on

$$\sum_{k=0}^n \frac{k}{n+1} \binom{n+1}{k} z^k (1-z)^{n+1-k} = z$$

Answer 1

We obtain \begin{align*} \color{blue}{\sum_{k=0}^n}&\color{blue}{\frac{k}{n}\binom{n}{k}z^k(1-z)^{n-k}}\\ &=\sum_{k=1}^n\frac{k}{n}\binom{n}{k}z^k(1-z)^{n-k}\tag{1}\\ &=\sum_{k=1}^n\binom{n-1}{k-1}z^k(1-z)^{n-k}\tag{2}\\ &=\sum_{k=0}^{n-1}\binom{n-1}{k}z^{k+1}(1-z)^{n-1-k}\tag{3}\\ &=z(z+(1-z))^{n-1}\tag{4}\\ &\,\,\color{blue}{=z}\tag{5} \end{align*} and the claim follows.

Comment:

• In (1) we skip the index with $k=0$.

• In (2) we use the binomial identity $\binom{p}{q}=\frac{p}{q}\binom{p-1}{q-1}$

• In (3) we shift the index $k$ by one and start with $k=0$. This is compensated by substituting each occurrence of $k$ with $k+1$.

• In (4) we factor out $z$ and apply the binomial theorem.

• In (5) we make a final simplification.