Use a factorial argument to show that $C(2n,n+1)+C(2n,n)=\frac{1}{2}C(2n+2,n+1)$

Question

I need some help, showing that the left hand side is equivalent to the right hand side. I tried but I get stuck, I am not sure if I am on the right path. Here is my attempt:

$C(2n,n+1) + C(2n,n)$

Equivalent to this

$\frac{(2n)!}{(n-1)!(n+1)!}+\frac{(2n)!}{n!n!}$

Now I add them together when I have a common denominator and I get

$\frac{(2n)!(n+n+1)}{(n)!(n+1)!}\frac{(2n)!(2n+1)}{(n)!(n+1)!} = \frac{(2n+1)!}{n!(n+1)!}$

But I do not know how to go from there. I don't know how to transform it to look like

$\frac{1}{2}C(2n+2,n+1)=\frac{(2n+2)!}{2(n+1)!(n+1)!}$

Any hint, tip, that will help me will be really useful.

Answer 1

Multiply and divide the answer you got by $2(n+1)$.

Answer 2

It’s usually a good idea to try to turn the more complicated expression into the simpler one. Here the more complicated one is

$$\frac{(2n+2)!}{2(n+1)!(n+1)!}\;,$$

and the simpler one is

$$\frac{(2n+1)!}{n!(n+1)!}\;.$$

Start with the numerator $(2n+2)!$; can it easily be represented in terms of $(2n+1)!$? Yes: $(2n+2)!=(2n+2)(2n+1)!$. What about the denominator? Can $2(n+1)!(n+1)!$ be written nicely in terms of $n!(n+1)!$? Again the answer is yes: $(n+1)!=(n+1)n!$, so

$$2(n+1)!(n+1)!=2(n+1)n!(n+1)!\;.$$

Thus,

$$\frac{(2n+2)!}{2(n+1)!(n+1)!}=\frac{(2n+2)(2n+1)!}{2(n+1)n!(n+1)!}\;,$$

and the way is clear to get $\dfrac{(2n+1)!}{n!(n+1)!}$ out of it.

Answer 3

$${\color{red}{1+1=2}}\quad|\cdot n\quad\iff\quad n+n=2n\quad|+1\quad\iff\quad n+n+1=2n+1\quad\iff$$

$$n+(n+1)=(2n+1)\quad|\cdot(n+1)\quad\iff\quad n(n+1)+(n+1)^2=(2n+1)(n+1)$$

$$\iff\qquad n(n+1)+(n+1)^2=(2n+1)\frac{2n+2}2\qquad\Bigg|\cdot\frac{(2n)!}{(n+1)!^2}\qquad\iff$$

$$\frac{(2n)!}{(n-1)!\cdot(n+1)!}+\frac{(2n)!}{n!^2}=\frac12\cdot\frac{(2n+2)!}{(n+1)!^2}\qquad\iff\qquad{\boxed{C_{2n}^{n\pm1}+C_{2n}^n=\frac12\cdot C_{2n+2}^{n+1}}}$$