Is it possible to show

Question

As part of a larger proof I need the following to be true in order to make it work,

$(a+1){b\choose a+1}x^{a+1}(1-x)^{b-a-1}=a{b\choose a}x^a(1-x)^{b-a}$

Can anyone give me a hand?

Answer 1

It’s just really a sequence of simplifications.

$$(a+1){b\choose a+1}x^{a+1}(1-x)^{b-a-1}=a{b\choose a}x^a(1-x)^{b-a}$$

$$\color{blue}{(a+1)}\frac{b!}{\color{blue}{(a+1)!}(b-(a+1))!}x^{a+1}(1-x)^{b-a-1}=\color{green}{a}\frac{b!}{\color{green}{a!}(b-a)!}x^a(1-x)^{b-a}$$

$$\frac{b!}{a!(b-a-1)!}\color{blue}{x^{a+1}}\color{green}{(1-x)^{b-a-1}}=\frac{b!}{(a-1)!(b-a)!}\color{blue}{x^a}\color{green}{(1-x)^{b-a}}$$

$$\frac{\color{blue}{b!}}{\color{green}{a!}\color{purple}{(b-a-1)!}}x=\frac{\color{blue}{b!}}{\color{green}{(a-1)!}\color{purple}{(b-a)!}}(1-x)$$

$$\frac{x}{a}=\frac{1-x}{b-a} \implies x(b-a)=a(1-x) \implies bx-ax=a-ax \implies bx = a$$

Answer 2

Unless $x=0$ or $x=1$, $$\frac{(a+1)b!}{(a+1)!(b-a-1)!}x^{a+1}(1-x)^{b-a-1}=\frac{ab!}{a!(b-a)!}x^a(1-x)^{b-a}$$

simplifies to

$$(b-a)x=a(1-x).$$