<s>[INST] Prove $\binom{2n+2}{n+1} = \binom{2n}{n+1} + 2\binom{2n}{n} + \binom{2n}{n-1}$ I need some help showing that these are equivalent. I made a couple attempts to get this right but so far the following work is as far as I've gotten.
Here is the question in its entirety:

Let n be a natural number. Give a combinatorial proof of the following:
  $\binom{2n+2}{n+1} = \binom{2n}{n+1} + 2\binom{2n}{n} + \binom{2n}{n-1}$

My first impression was that I could use the $\binom{n}{k} = \binom{n}{n-k}$ identity to make the term "$\binom{2n}{n-1}$" equal "$\binom{2n}{n+1}$" so I could simplify the equation a bit. I now have:
$\binom{2n+2}{n+1} = 2\binom{2n}{n+1} + 2\binom{2n}{n} = 2[\binom{2n}{n+1} + \binom{2n}{n}]$
Afterwards i figured I could use pascals identity $\binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}$ to get rid of yet more terms resulting in:
$\binom{2n +2}{n+1} = 2\binom{2n+1}{n+1}$
This is where I realize I made a pretty big mistake and got stuck, I'm fairly sure I made a maths error somewhere but I am unsure where after my third attempt at this.
 [/INST] The question asks for combinatorial proof. But going by your way, ie using Pascal's identity it can be done as follows:
$$\binom{2n}{n} + \binom{2n}{ n-1} = \binom{2n+1}{n}$$
And $$\binom{2n}{n} + \binom{2n}{n+1} = \binom{2n+1}{n+1}$$
so adding both the equations,
$$\begin{align}
\binom{2n}{ n-1} +2\binom{2n}{n}+ \binom{2n+1}{n} &= \binom{2n+1}{n} + \binom{2n+1}{n+1}\\ 
&= \binom{2n+2}{n+1}
\end{align}$$
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