Observe that \[\frac{1}{1}+\frac{1}{3}=\frac{4}{3}, \;\; 4^{2}+3^{2}=5^{2},\] \[\frac{1}{3}+\frac{1}{5}=\frac{8}{15}, \;\; 8^{2}+{15}^{2}={17}^{2},\] \[\frac{1}{5}+\frac{1}{7}=\frac{12}{35}, \;\;{12}^{2}+{35}^{2}={37}^{2}.\] State and prove a generalization suggested by these examples.
Problem
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Tags: Miscellaneous Problems
Peter
22.07.2007 06:22
Robert Gerbicz wrote: For all positive integer n the following is true: $ \frac{1}{2*n-1}+\frac{1}{2*n+1}=\frac{a}{b}$ where $ {a}^{2}+{b}^{2}={((2*n-1)*(2*n+1)+2)}^{2}$ Proof: $ \frac{1}{2*n-1}+\frac{1}{2*n+1}=\frac{4*n}{4*{n}^{2}-1}$ so $ a=4*n$ and $ b=4*{n}^{2}-1$ is good choice and $ {a}^{2}+{b}^{2}={((2*n-1)*(2*n+1)+2)}^{2}$ is also true, because $ {((2*n-1)*(2*n+1)+2)}^{2}-{b}^{2}={(4*{n}^{2}+1)}^{2}-{(4*{n}^{2}-1)}^{2}= 2*8*{n}^{2}=16*{n}^{2}={a}^{2}$.
manjil
17.10.2008 04:27
Am I missing something or the problem is as easy as it looks?