How to show $\sum_{i=1}^{n} \binom{i}{2}=\binom{n+1}{3}$? 
Show that $\,\displaystyle\sum_{i=1}^{n} \binom{i}{2}=\binom{n+1}{3}$.

I'm thinking right now (though not getting anywhere with it) that I want to expand out the summation portion to $i!/2!(i-2)!$ and simplify from there? Not sure if that will help, not to mention if I put $1$ in for $i$ I get $1/-2$ which I don't think is right. 
Anyone care to shed some light on the subject? 
Thanks.
 A: The question seems to be about how to calculate $\binom{1}{2}$. In general, for $n,k \geq 0$, the binomial coefficient can be defined to be the coefficient of $x^k$ in the expansion of $(1+x)^n$. In other words, $\binom{1}{2}$ is the coefficient of $x^2$ in $(1+x)^1$. But this coefficient is 0. Hence $\binom{1}{2} = 0$, which is why this doesn't affect the validity of the proofs in the other answers.
As another way to think about it, $\binom{1}{2}$ counts the number of ways to choose $2$ objects from a set of $1$; there are no such ways. In general, $\binom{n}{k} = 0$ for $k > n \geq 0$.
A: What we shall show is that
$$
\sum_{i=2}^n\binom{i}{2}=\binom{n+1}{3},
$$
using a combinatorial argument:
The right-hand side is the number of the ways to choose three numbers among $\{1,2,\ldots,n,n+1\}$.
This is equal to the sum of $P_i$'s, $i=2,\ldots,n$, where $P_i$ is the number of ways to choose three numbers among  $\{1,2,\ldots,n,n+1\}$ with the largest one being $i+1$.
 But $P_i$ is the number of ways to choose two numbers among $\{1,2,\ldots,i\}$, which is equal to $\binom{i}{2}$. Hence
$$
\binom{n+1}{3}=\binom{2}{2}+\binom{3}{2}+\cdots+\binom{n}{2}.
$$
A: Well, if $n-1 =2k$ $$\displaystyle\sum_{i=1}^{n} \binom{i}{2} = \displaystyle\sum_{i=1}^{n}\frac{i\cdot (i-1)}{2}=\frac{1\cdot 0}{2}+\frac{2\cdot 1}{2}+\frac{3\cdot 2}{2}+\frac{4\cdot 3}{2} + \ldots +\frac{n\cdot (n-1)}{2} = \frac{2\cdot 1+3\cdot 2+4\cdot 3+ \ldots +n\cdot (n-1)}{2} = \frac{2^2\cdot2+4^2\cdot2+6^2\cdot2+\ldots +(n-1)^2\cdot2}{2}=2^2+4^2+6^2+\ldots+(n -1)^2=4(1+2^2+3^2+\ldots+k^2)=\frac{4k\cdot(k+1)\cdot(2k+1)}{6}=\frac{4\frac{n-1}{2}\cdot\frac{n+1}{2}\cdot n}{6} = \frac{(n+1)\cdot n\cdot(n-1)}{3!}=\binom{n+1}{3}$$
Similarly if $n-1 =2k +1$ you have $$\displaystyle\sum_{i=1}^{n} \binom{i}{2} = \displaystyle\sum_{i=1}^{n}\frac{i\cdot (i-1)}{2}=\frac{1\cdot 0}{2}+\frac{2\cdot 1}{2}+\frac{3\cdot 2}{2}+\frac{4\cdot 3}{2} + \ldots +\frac{n\cdot (n-1)}{2} = \frac{2\cdot 1+3\cdot 2+4\cdot 3+ \ldots +n\cdot (n-1)}{2} = \frac{2^2\cdot2+4^2\cdot2+6^2\cdot2+\ldots +(n-2)^2\cdot2+n\cdot(n-1)}{2}=2^2+4^2+6^2+\ldots+(n-2)^2+n\cdot (n-1)=4(1+2^2+3^2+4^2+\ldots+(n-2)^2)+\frac{n\cdot(n-1)}{2}=\frac{4k\cdot(k+1)\cdot(2k+1)}{6}+\frac{n\cdot(n-1)}{2}=\frac{4\frac{n-2}{2}\cdot\frac{n}{2}\cdot (n-1)}{6}+\frac{n\cdot(n-1)}{2}=\frac{(n-2)\cdot n\cdot(n-1)}{3!}+\frac{n\cdot(n-1)}{2}=\frac{n\cdot(n-1)\cdot(n-2)+3n\cdot(n-1)}{3!}=\frac{n\cdot(n-1)\cdot(n-2+3)}{3!}=\frac{n\cdot(n-1)\cdot(n+1)}{3!}=\binom{n+1}{3}$$
A: Hint.  The basic Pascal's triangle identity is
$$\binom{i+1}{k}=\binom{i}{k}+\binom{i}{k-1}\ .$$
Taking $k=3$ and rearranging,
$$\binom{i}{2}=\binom{i+1}{3}-\binom{i}{3}\ .$$
Therefore the sum is
$$\sum_{i=1}^n \left[\binom{i+1}{3}-\binom{i}{3}\right]\ ,$$
which is a telescoping sum.
A: $$\sum_{i=2}^n\binom{i}{2}=\binom{2}{2}+\binom{3}{2}+\binom{4}{2}+\binom{5}{2}+\cdots+\binom{n}{2}$$
Since $\binom22=\binom33$
$$\sum_{i=2}^n\binom{i}{2}=\binom{3}{3}+\binom{3}{2}+\binom{4}{2}+\binom{5}{2}+\cdots+\binom{n}{2}$$
Pascal's Triangle Identity states that
$$\binom{n-1}k+\binom{n-1}{k-1}=\binom{n}{k}$$
So,
$$\sum_{i=2}^n\binom{i}{2}=\color{red}{\binom{3}{3}+\binom{3}{2}}+\binom{4}{2}+\binom{5}{2}+\cdots+\binom{n}{2}$$
$$\sum_{i=2}^n\binom{i}{2}=\color{red}{\binom{4}{3}}+\binom{4}{2}+\binom{5}{2}+\cdots+\binom{n}{2}$$
$$\sum_{i=2}^n\binom{i}{2}=\color{green}{\binom{4}{3}+\binom{4}{2}}+\binom{5}{2}+\cdots+\binom{n}{2}$$
$$\sum_{i=2}^n\binom{i}{2}=\color{green}{\binom{5}{3}}+\binom{5}{2}+\cdots+\binom{n}{2}$$
So on and so forth, until we have
$$\sum_{i=2}^n\binom{i}{2}=\binom{n}{3}+\binom{n}{2}=\binom{n+1}{3}$$
A: $\newcommand{\+}{^{\dagger}}
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$\ds{\sum_{i = 1}^{n}{i \choose 2} = {n + 1 \choose 3}:\ {\large ?}}$

\begin{align}
\color{#00f}{\large\sum_{k = 1}^{n}{k \choose 2}}&=\sum_{k = 1}^{n}
\int_{\verts{z}
= 1}
{\pars{1 + z}^{k} \over z^{3}}{\dd z \over 2\pi\ic}
=\int_{\verts{z} = 1}{\dd z \over 2\pi\ic}\,{1 \over z^{3}}
\sum_{k = 1}^{n}\pars{1 + z}^{k}
\\[3mm]&=\int_{\verts{z} = 1}{\dd z \over 2\pi\ic}\,{1 \over z^{3}}\,
{\pars{1 + z}\bracks{\pars{1 +z}^{n} - 1} \over \pars{1 +z} - 1}
\\[3mm]&=\underbrace{%
\int_{\verts{z} = 1}{\dd z \over 2\pi\ic}\,{\pars{1 + z}^{n + 1} \over z^{4}}}
_{\ds{=\ {n + 1 \choose 3}}}\
-\
\overbrace{\int_{\verts{z} = 1}{\dd z \over 2\pi\ic}\,{1 + z \over z^{4}}}
^{\ds{=\ 0}}
=\color{#00f}{\large{n + 1 \choose 3}}
\end{align}

A: The process of choosing all combinations of three from $n+1$ can be thought about like this:
If we define $e_i$ to be the ith element,


*

*Choose $e_1$ and then choose 2 other elements from the rest. These would be all combinations that have $e_1$ as a choice.

*Choose $e_2$ as an element and then choose 2 other elements from the n-1 after $e_2$. These would be all combinations that have $e_2$ as a choice, but not $e_1$, as it  was already counted.

*Choose $e_3$ as an element and then choose 2 others from the n-2 after $e_3$. These would be all combinations that have $e_3$ as a choice, excluding $e_1$ and $e_2$ since they were already counted.


...
