Find the Sum $1\cdot2+2\cdot3+\cdots + (n-1)\cdot n$ Find the sum $$1\cdot2 + 2\cdot3 + \cdot \cdot \cdot + (n-1)\cdot n.$$
This is related to the binomial theorem.  My guess is we use the combination formula . . .
$C(n, k) = n!/k!\cdot(n-k)!$
so . . . for the first term $2 = C(2,1) = 2/1!(2-1)! = 2$
but I can't figure out the second term $3 \cdot 2 = 6$ . . . 
$C(3,2) = 3$ and $C(3,1) = 3$ I can't get it to be 6.
Right now i have something like . . .
$$ C(2,1) + C(3,2) + \cdot \cdot \cdot + C(n, n-1) $$
The 2nd term doesn't seem to equal 6.
What should I do? 
 A: This doesn't solve it using the binomial theorem, but this is one way to do it:
The general term is given by $a_n =n(n+1) = n^2 + n$.
So the sum to $n$ terms: $$ \sum_{i=1}^{n} a_i = \sum_{i=1}^{n} i^2 + \sum_{i=1}^{n} i$$
As you probably may know,
$\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}$
$\sum_{i=1}^{n} i = \frac{n(n+1)}{2}$
A: HINT:
$$\begin{align*}
1\cdot2+2\cdot3+\ldots+n(n+1)&=2\binom22+2\binom32+\ldots+2\binom{n+1}2\\
&=2\left(\binom22+\binom32+\ldots+\binom{n+1}2\right)\;,
\end{align*}$$
and you can apply the Christmas stocking (or hockey stick) identity, if you know it. (If not, you might also want to look at this demonstration.)
A: As I have been directed to teach how to fish... this is a bit clunky, but works.
Define rising factorial powers:
$$
x^{\overline{m}} = \prod_{0 \le k < m} (x + k) = x (x + 1) \ldots (x + m - 1)
$$
Prove by induction over $n$ that:
$$
\sum_{0 \le k \le n} k^{\overline{m}} = \frac{n^{\overline{m + 1}}}{m + 1}
$$
When $n = 0$, it reduces to $0 = 0$.
Assume the formula is valid for $n$, and:
$$
\begin{align*}
\sum_{0 \le k \le n + 1} k^{\overline{m}}
  &= \sum_{0 \le n} k^{\overline{m}} + (n + 1)^{\overline{m}} \\
  &= \frac{n^{\overline{m + 1}}}{m + 1} + (n + 1)^{\overline{m}} \\
  &= \frac{n \cdot (n + 1)^{\overline{m}} + (m + 1) (n + 1)^{\overline{m}}}
          {m + 1} \\
  &= \frac{(n + m + 1) \cdot (n + 1)^{\overline{m}}}{m + 1} \\
  &= \frac{(n + 1)^{\overline{m + 1}}}{m + 1}
\end{align*}
$$
By induction, it is valid for all $n$.
Defining falling factorial powers:
$$
x^{\underline{m}} = \prod_{0 \le k < m} (x - k) = x (x - 1) \ldots (x - m + 1)
$$
you get a similar formula for the sum:
$$
\sum_{0 \le k \le n} k^{\underline{m}}
$$
You can see that $x^{\overline{m}}$ (respectively $x^{\underline{m}}$) is a monic polynomial of degree $m$, so any integral power of $x$ can be expressed as a combination of appropiate factorial powers, and so sums of polynomials in $k$ can also be computed with some work.
By the way, the binomial coefficient:
$$
\binom{\alpha}{k} = \frac{\alpha^{\underline{k}}}{k!}
$$
A: Check for a relationship between $n (n + 1) (n + 2)$ and your sum by checking small cases and prove it by induction.
