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?
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.