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

Printable View

- Sep 12th 2010, 10:10 PMtaylor1234Induction Method
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 - Sep 12th 2010, 11:00 PMProve It
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.