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.
 A: Consider picking $n+1$ objects from a set of $2n+2$ objects, where two of these are "special", and the other $2n$ are unspecial.
One way is simple : to pick $n+1$ objects from $2n+2$ objects, the number of ways are $\binom {2n+2}{n+1}$ clearly. 
The other way. We split in three cases.
One can pick none of these special objects, and pick $n+1$ from the  $2n$ unspecial objects. This is done in $\binom{2n}{n+1}$ ways.
Next, pick any one special object in two ways, and pick the rest of the $n$ objects from the unspecial ones. This is done in $2 \times \binom{2n}{n}$ ways.
Finally, one can pick both special objects and pick the rest of the $n-1$ objects from the unspecial ones in $\binom{2n}{n-1}$ ways.
Since the above ways are disjoint (they contain different number of special objects) and cover all possible cases, adding these gives you your result.
A: \begin{eqnarray*}
 \binom{2n+2}{n+1}= \color{red}{\binom{2n+1}{n+1}}+\color{blue}{\binom{2n+1}{n}}=\color{red}{\binom{2n}{n+1}+\binom{2n}{n}}+\color{blue}{\binom{2n}{n}+\underbrace{\binom{2n}{n-1}}_{\binom{2n}{n-1}=\binom{2n}{n+1}}}=2 \left(\binom{2n}{n+1}+\binom{2n}{n}\right)
\end{eqnarray*}
A: 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,
$$\begin{align}
\binom{2n}{ n-1} +2\binom{2n}{n}+ \binom{2n+1}{n} &= \binom{2n+1}{n} + \binom{2n+1}{n+1}\\ 
&= \binom{2n+2}{n+1}
\end{align}$$
A: Left hand side is the coefficient of $x^{n+1}$ in $(1+x)^{2n+2}$
Now,
$$(1+x)^{2n+2}  = (1+x)^{2n}(1+x)^2 = (1+x)^{2n}(1+2x+x^2)$$
The coefficient of $x^{n+1}$ on the right hand side is 
$$\binom{2n}{n+1} + 2\binom{2n}{n} + \binom{2n}{n-1}$$
It follows that
$$\binom{2n+2}{n+1} = \binom{2n}{n+1} + 2\binom{2n}{n} + \binom{2n}{n-1}$$
