What is the following sum?
(1/1*2) + (1/2*3) + (1/3*4) + . . . + (1/(n-1)n)
Experiment, Conjucture the value, and then prove by induciton
/ means division, ex. 1/2=.5
I assume that this is
$\displaystyle \frac{1}{1\cdot 2} + \frac{1}{2\cdot 3} + \frac{1}{3\cdot 4} + \dots + \frac{1}{(n - 1)n}$.
Notice that you can write the $\displaystyle k^{\textrm{th}}$ term as
$\displaystyle \frac{1}{k(k + 1)} = \frac{1}{k} - \frac{1}{k + 1}$.
So that means we can rewrite the sum as
$\displaystyle \frac{1}{1\cdot 2} + \frac{1}{2\cdot 3} + \frac{1}{3\cdot 4} + \dots + \frac{1}{(n - 1)n} = \left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \dots + \left(\frac{1}{n - 1} - \frac{1}{n}\right)$
$\displaystyle = 1 - \frac{1}{n}$, since all but the first and last term cancel.