# Explicit form

• Nov 22nd 2007, 06:11 AM
james_bond
Explicit form
$1*2+2*3+3*4+\cdots +(n-1)n=?$
• Nov 22nd 2007, 06:54 AM
math sucks
$1\times2+2\times3+3\times4+\ldots+(n-1)n$
$=2+6+12+20+\ldots+(n^2-n)$
$=2\left(1+3+6+10+\ldots+\frac{n^2-n}{2}\right)$
Inside the parenthesis is the sum of the first $n-1$ triangular numbers. This can be given by the formula,
$\frac{(n-1)(n)(n+1)}{6}$
Then,
$2\left(1+3+6+10+\ldots+\frac{n^2-n}{2}\right)
=\frac{(n-1)(n)(n+1)}{3}$
• Nov 22nd 2007, 06:59 AM
Soroban
Hello, James!

Quote:

Find the explicit form for: . $1\cdot2+2\cdot3+3\cdot4 + \cdots +(n-1)n$
It helps if you know some summation formulas . . .

We have: . $\sum^n_{k=1}k(k-1) \;=\;\sum^n_{k=1}(k^2 - k)$

. . $=\qquad\quad\sum^n_{k=1} k^2 \quad-\qquad \sum^n_{k=1} k$

. . $=\;\overbrace{\frac{k(k+1)(2k+1)}{6}} - \overbrace{\frac{k(k+1)}{2}}$

Factor: . $\frac{k(k+1)}{6}\cdot\left[(2k+1) - 3\right] \;=\;\frac{k(k+1)}{6}\cdot(2k-2) \;=\;\frac{k(k+1)}{6}\cdot2(k-1)$

Therefore: . $\boxed{\frac{(n-1)n(n+1)}{3}}$