Using the Binomial Identity, prove that ${n\choose k}+2{n\choose k+1}+{n\choose k+2}={n+2\choose k+2}$ Using the Binomial Identity, prove that: $${n\choose k}+2{n\choose k+1}+{n\choose k+2}={n+2\choose k+2}$$Because this is in the form of a Binomial Coefficient, I can break down the LHS further:$$\left({n\choose k}+{n\choose k+1}\right)+\left({n\choose k+1}+{n\choose k+2}\right)$$From here, however, an answer key that I'm using immediately jumps to:$$={n+1\choose k+1}+{n+1\choose k+2}$$Which then jumps to:$$={n+2\choose k+2}$$But I don't know how either of those last two steps were reached after I break down the LHS. Could someone clarify these steps for me?
 A: These steps come from

$$
{n\choose p}+{n\choose p+1}={n+1\choose p+1}. \tag1
$$

Then, just apply $(1)$ the first time with $n:=n, \,p:=k$ and the second time with $n:=n, \,p:=k+1$ and the third time with $n:=n+1, \,p:=k+1$.
Can you take it from here?
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
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 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

With $\ds{\quad a_{k} = a_{k + 2} = 1\,,\quad a_{k + 1} = 2}$:


\begin{align}
\color{#f00}{\sum_{j = k}^{k + 2}a_{j}{n \choose j}} & =
\sum_{j = k}^{k + 2}a_{j}\oint_{\verts{z} = 1}
{\pars{1 + z}^{n} \over z^{\ j + 1}}\,{\dd z \over 2\pi\ic} =
\oint_{\verts{z} = 1}
{\pars{1 + z}^{n} \over z}\,\,\sum_{j = k}^{k + 2}{a_{j} \over z^{\ j}}
\,{\dd z \over 2\pi\ic}
\\[4mm] & =
\oint_{\verts{z} = 1}{\pars{1 + z}^{n} \over z}
\pars{{1 \over z^{k}} + {2 \over z^{k + 1}} + {1 \over z^{k + 2}}}
\,{\dd z \over 2\pi\ic}
\\[4mm] & =
\oint_{\verts{z} = 1}{\pars{1 + z}^{n} \over z^{k + 3}}
\pars{z^{2} + 2z + 1}\,{\dd z \over 2\pi\ic} =
\oint_{\verts{z} = 1}{\pars{1 + z}^{n + 2} \over z^{k + 3}}
\,{\dd z \over 2\pi\ic} = \color{#f00}{n + 2 \choose k + 2}
\end{align}
