Proving combinatorial identities

Question

Prove $\displaystyle\sum_{k=1}^n kx^k{x\choose k}=nx(1+x) ^{n-1}$

This question can be solved easily (by taking the derivative of the binomial theorem formula), if there was an $\binom{n}{k}$ instead of $\binom{x}{k}$. I mean, the presence of $\binom{x}{k}$ seems a bit fishy. so, is this question correct? If yes, could you give a hint on how to solve it?

Source: A first course in probability theory, Sheldon Ross (9th edition)

Answer 1

It's probably just a typo.

this question can be solved easily (by taking the derivative of the binomial theorem formula)

and multiplying both sides by $x$ you get

$nx(1+x)^{n-1} = \sum\limits_{k=1}^n k x^k{n\choose k}$.

If what is typed were also true we'd have

$ \sum\limits_{k=1}^n k x^k{n\choose k} = \sum\limits_{k=1}^n k x^k{x\choose k}$

.... for every possible value of $n$ and $x$. That's surely nuts!

A simple counter-example would be $x=1$.

The left hand side $n(1+1)^{n-1} = \sum\limits_{k=1}^n k {n\choose k}= n2^{n-1}$ for all $n$. (Mildly interesting... so $5*2^4 = 80$ is equal to ${5\choose 1}+2{5\choose 2} + 3{5\choose 3}+4{5\choose 4}+5{5\choose 5}=5+20 + 30 + 20+5$.... cool.... I guess.....That's a funny side shift I never noticed... well, I'll play with that later... but I digress)

But the right hand side is $\sum\limits_{k=1}^n k x^k{x\choose k}= 1*1^1{1\choose k} + \sum_{k=2}^n k1^k{1\choose k} = 1 + \sum_{k=1}^n 0 = 1$.

Obviously $n2^{n-1} =1$ is not a universal identity for all $n$ so the statement is false.

It's probably just a typo.

Answer 2

Keep in mind $\binom{x}{k}=0 \ \forall k>x$. If $x \in \mathbb{R}$, there's a nice upperbound by taking $\binom{x}{k} \leq \frac{x^k}{k!}$ and $n \to \infty$: $$ \sum_{k=1}^{n}k \frac{x^{2k}}{k!} \to_n x^2 e^{x^2} $$