mathematical induction
- Thread starter Safia
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Dr.Peterson
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- Nov 12, 2017
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This appears to say:
Prove by mathematical induction that for all positive integers n:
\(\displaystyle \displaystyle \frac{1}{1\cdot 3} + \frac{1}{3\cdot 5} + \frac{1}{5\cdot 7} + ... + \frac{1}{(2n-1)(2n+1)} = \frac{n}{2n+1}\)
\(\displaystyle \displaystyle \frac{1}{1\cdot 3} + \frac{1}{3\cdot 5} + \frac{1}{5\cdot 7} + ... + \frac{1}{(2n-1)(2n+1)} = \frac{n}{2n+1}\)
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