# Really confusing?

• Apr 27th 2009, 06:40 PM
yoman360
Really confusing?
use (1/x) - (1/x+1) to find the following sum.
(1/1*2) + (1/2*3) + (1/3*4) ..... + (1/99*100)

show/explain how to get the answer

the answer is 99/100 if that helps
• Apr 27th 2009, 07:33 PM
Chris L T521
Quote:

Originally Posted by yoman360
use (1/x) - (1/x+1) to find the following sum.
(1/1*2) + (1/2*3) + (1/3*4) ..... + (1/99*100)

show/explain how to get the answer

the answer is 99/100 if that helps

When you expand the sum, something very nice will happen...

$\displaystyle \sum_{x=1}^{99}\left(\frac{1}{x}-\frac{1}{x+1}\right)=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\dots+\left(\frac{1}{99}-\frac{1}{100}\right)$ $\displaystyle =1-\frac{1}{100}=\boxed{\frac{99}{100}}$

Does this make sense?
• Apr 27th 2009, 08:08 PM
Prove It
Quote:

Originally Posted by Chris L T521
When you expand the sum, something very nice will happen...

$\displaystyle \sum_{x=1}^{99}{\left(\frac{1}{x}-\frac{1}{x+1}\right)}=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\dots+\left(\frac{1}{99}-\frac{1}{100}\right)$ $\displaystyle =1-\frac{1}{100}=\boxed{\frac{99}{100}}$

Does this make sense?

Note that your original series is

$\displaystyle \frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \frac{1}{3 \times 4} + \dots + \frac{1}{99 \times 100} = \sum_{i = 1}^{99}{\frac{1}{x(x + 1)}}$

Also note that $\displaystyle \frac{1}{x} - \frac{1}{x + 1} = \frac{1}{x}\times\frac{x + 1}{x + 1} - \frac{1}{x + 1} \times \frac{x}{x}$

$\displaystyle = \frac{x + 1}{x(x + 1)} - \frac{x}{x( x + 1)}$

$\displaystyle = \frac{x + 1 - x}{x(x + 1)}$

$\displaystyle = \frac{1}{x(x + 1)}$.

So $\displaystyle \sum_{i = 1}^{99}{\frac{1}{x(x + 1)}} = \sum_{i = 1}^{99}{\left(\frac{1}{x} - \frac{1}{x + 1}\right)}$.

Then refer to Chris' post.
• Apr 28th 2009, 03:42 PM
yoman360
Quote:

Originally Posted by Chris L T521
When you expand the sum, something very nice will happen...

$\displaystyle \sum_{x=1}^{99}\left(\frac{1}{x}-\frac{1}{x+1}\right)=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\dots+\left(\frac{1}{99}-\frac{1}{100}\right)$ $\displaystyle =1-\frac{1}{100}=\boxed{\frac{99}{100}}$

Does this make sense?

Wait How did you get 1-(1/100)?
• Apr 28th 2009, 05:07 PM
Prove It
Quote:

Originally Posted by yoman360
Wait How did you get 1-(1/100)?

$\displaystyle -\frac{1}{2} + \frac{1}{2} = 0$

$\displaystyle -\frac{1}{3} + \frac{1}{3} = 0$

$\displaystyle -\frac{1}{4} + \frac{1}{4} = 0$ etc.

So everything cancels except $\displaystyle 1 - \frac{1}{100}$.
• Apr 28th 2009, 05:32 PM
yoman360
Quote:

Originally Posted by Prove It
$\displaystyle -\frac{1}{2} + \frac{1}{2} = 0$

$\displaystyle -\frac{1}{3} + \frac{1}{3} = 0$

$\displaystyle -\frac{1}{4} + \frac{1}{4} = 0$ etc.

So everything cancels except $\displaystyle 1 - \frac{1}{100}$.

oh thanks man. I don't know why I didn't notice