How to prove the binomial identity $\binom{n}{r - 1} + 2\binom{n}{r} + \binom{n}{r + 1} = \binom{n + 2}{r + 1}$?

Question

Prove that $\dbinom{n}{r - 1} + 2\dbinom{n}{r} + \dbinom{n}{r + 1} = \dbinom{n + 2}{r + 1}$.

I've tried manipulating both sides with $\dbinom{n}{r}$ factored out but I'm stuck. Any help?

Answer 1

Hint: Apply Pascal's Identity $$\binom{m}{k} = \binom{m - 1}{k} + \binom{m - 1}{k - 1}$$ to the right-hand side, then apply it to each of the resulting terms.

Answer 2

Using Vandermonde's Identity, $$\begin{align} \binom n{r-1}+2\binom nr+\binom n{r+1} &=\sum_{i=0}^2 \binom 2{2-i} \binom n{r-1+i}\\ &=\binom {n+2}{r+1}\end{align}$$

Answer 3

Take a pile of $n$ green beads, and $2$ blue beads.   Now the way to select $r+1$ of those $n+2$ beads can be counted as:

$$\binom 22\binom{n}{r-1}+\binom 21\binom{n}r+\binom 20\binom n{r+1}$$

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But wait, $\dbinom {n+2}{r+1}$ is also the count of ways to select $r+1$ items from $n+2$.   ....

Answer 4

Consider the coefficient of $x^{r+1}$ in $(1+x)^{n+2}$. This is $\binom{n+2}{r+1}$. On the other hand, \begin{align*} (1+x)^{n+2} &= (1+x)^{n}(1+x)^2 \\ &= (1+x)^n(1+2x+x^2) \end{align*} The coefficient of $x^{r+1}$ on the right hand side is $$\binom{n}{r-1} + 2\binom{n}{r} + \binom{n}{r+1}$$