Proving ${n + 2 \choose m + 1} = {n \choose m + 1} + 2{n \choose m} + {n \choose m - 1}$ using the Binomial Theorem

Question

I want to prove the following identity using the Binomial Theorem: $${n + 2 \choose m + 1} = {n \choose m + 1} + 2{n \choose m} + {n \choose m - 1}$$

I have a combinatorial solution: Suppose we want to choose $m + 1$ people among $n + 2$ people where 2 of these people are children. This can be done in ${n + 2 \choose m}$ ways. Now to prove the identity we count this differently by looking at the different cases: pick no children, pick 1 child, pick 2 children.

But I want to prove this using the Binomial Theorem but I have no idea how this can be done. I only know that ${n + 2 \choose m + 1}$ is the coefficient of $x^{m + 1}y^{n + 1 - m}$ in $(x + y)^{n + 2}$ but I haven't been able to do anything with this. I tried different values of $x$ and $y$ and taking derivatives but I couldn't solve the problem.

Hints would be appreciated. Thanks in advance.

Edit: This question is pretty much the same as mine, however there are no solutions using the Binomial theorem there.

Answer 1

Expand $(x+y)^n$ and $(x+y)^2$ separately, then multiply the two expansions together.

Answer 2

Hint: let $[x^k]f(x)$ denote the $x^k$ coefficient in $f(x)$ so$$\binom{n+2}{m+1}=[x^{m+1}](1+x)^{n+2}=[x^{m+1}](1+x)^n(1+2x+x^2).$$(In other words, what @Arthur said, but we only need the case $y=1$.)