How can I simplify $1\times 2 + 2 \times 3 + .. + (n-1) \times n$ progression? I have a progression that goes like this:
$$1\times 2 + 2 \times 3 + .. + (n-1) \times n$$
Is there a way I can simplify it?
 A: Hint. You have
$$
(n^3-n)-\left((n-1)^3-(n-1)\right)=3\times(n-1)\times n
$$ and the sum is telescoping.
A: You are considering $$S=\sum_{i=1}^{n-1}i\,(i+1)=\sum_{i=1}^{n-1}i^2+\sum_{i=1}^{n-1}i$$ which are quite standard summations. For the first one $$\sum_{i=1}^{n-1}i^2=\frac{1}{6} (n-1) n (2 n-1)$$ and for the second $$\sum_{i=1}^{n-1}i=\frac{1}{2} (n-1) n$$ Add them together and simplify (noticing some nice factorizations).
A: Yes.
\begin{align*}
1\cdot 2 + 2\cdot 3 + \dotsb + (n-1)\cdot n
&= 1\cdot(1+1) + 2\cdot(2+1) + \dotsb + (n-1)\cdot((n-1)+1) \\
&= 1^2 + 1 + 2^2 + 2 + \dotsb + (n-1)^2 + (n-1) \\
&= (1^2+2^2+\dotsb+(n-1)^2) + (1+2+\dotsb+(n-1))
\end{align*}
Both of those sums have well-known closed forms; see here, for example.
A: You have $T_r=r(r+1)$ which you can write as
$$T_r=\frac{r(r+1)\{(r+2)-(r-1)\}}{(r+2)-(r-1)}$$
$$T_r=\frac{1}{3}\{r(r+1)(r+2)-(r-1)r(r+1)\}$$
And there you have your telescoping series. On adding up all terms you will find 
$$S_{n-1}=\frac{1}{3}(n-1)n(n+1)$$
