Show that $\dbinom{2n}{n} + \dbinom{2n}{n-1} = \frac{1}{2} \dbinom{2n+2}{n+1}$
By induction, suppose that for some n its true, $\dbinom{2n}{n} + \dbinom{2n}{n-1} = \dbinom{2n+2}{n+1}$ by theorem, but, I don't know how to prove that $\dbinom{2n+2}{n+1} = \frac{1}{2} \dbinom{2n+4}{n+2}$
On
Say you have 2 red balls and other white. On how many ways you can choose $n+1$ of them? One answer is ${2n+2\choose n+1}$.
And second answer:
If both red are choosen then we must choose $n-1$ balls beetwen $2n$ balls, so that is $${2n\choose n-1}$$
If one red is choosen then we must choose $n$ balls beetwen $2n$ balls and one red, that is two ways, so that is $$2{2n\choose n}$$
If no red is choosen then we must choose $n+1$ balls beetwen $2n$ balls, so that is $${2n\choose n+1} = {2n\choose n-1}$$
Sum that and you are done.
$$\begin{align} {2n\choose n}+{2n\choose n-1}&=\frac{(2n)!}{n!n!}+\frac{(2n)!}{(n-1)!(n+1)!} \\ &= \frac{(2n)!(n+1)}{n!(n+1)!}+\frac{(2n)!n}{n!(n+1)!} \\ &= \frac{(2n)!(n+1+n)}{n!(n+1)!} \\ &= \frac{(2n+1)!}{n!(n+1)!}\tag{1} \end{align}$$ And $$\begin{align} \frac 12{2n+2\choose n+1}&=\frac 12\frac{(2n+2)!}{(n+1)!(n+1)!} \\ &= \frac 12\frac{(2n+2)(2n+1)!}{(n+1)n!(n+1)!} \\ &=\frac{2n+2}{2(n+1)}\cdot\frac{(2n+1)!}{n!(n+1)!} \\ &= \frac{(2n+1)!}{n!(n+1)!}\tag{2} \end{align}$$ Finally, combine $(1)$ and $(2)$. You can also use Pascal's identity to immediately obtain the result in $(1)$.