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 08-04-2007 #27 Sillysidley   Join Date: Oct 2006 Posts: 822 Eh, close enough The first one uses a telescoping series (basically where a bunch of things cancel out) and the second has a formula (and just to get better at proofs, I wanna prove it) Show more: We want all the terms except the first and last to cancel. So, we express it as: $\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}...+\frac{1}{499}-\frac{1}{500}=1-\frac{1}{500}=\frac{499}{500}$ 2nd:Show more: There's a formula that says the sum of n numbers of this kind sum up to $\frac{n}{n+1}$. In other words, $\frac{1}{1*2}+\frac{1}{2*3}+...+\frac{1}{n*(n+1)}= \frac{n}{n+1}$We can then easily see the answer is $\frac{499}{500}$. Proof: Easy enough to prove by induction: For n=1, the LHS=RHS=$\frac{1}{1*2}$. Let's assume it holds for n. Then, for n+1; $\frac{1}{1*2}+\frac{1}{2*3}+...+\frac{1}{(n+1)(n+2 )}=\frac{n}{n+1}+\frac{1}{(n+1)(n+2)}= \frac{n(n+2)}{(n+1)(n+2)}+\frac{1}{(n+1)(n+2)} =\frac{n^2+2n+1}{(n+1)(n+2)}=\frac{n+1}{n+2}$ And we're finally done __________________ . Last edited by Sillysidley; 08-04-2007 at 05:06 PM..