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PFStudent
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Homework Statement
1.1.2
Write,
[tex]
{\frac{1}{1\cdot2}} + {\frac{1}{2\cdot3}} + {\frac{1}{3\cdot4}} + . . . + {\frac{1}{99\cdot100}}
[/tex]
as a fraction in lowest terms.
The Attempt at a Solution
Rewriting the problem as a summation,
[tex]
{{\sum_{i = 1}^{n}}{\frac{1}{i(i+1)}}} = {\frac{1}{1\cdot2}} + {\frac{1}{2\cdot3}} + {\frac{1}{3\cdot4}} + . . . + {\frac{1}{n\cdot(n+1)}}
[/tex]
Then considering the first few terms,
[tex]
{\frac{1}{1\cdot2}} = \frac{1}{2}
[/tex]
[tex]
{\frac{1}{1\cdot2}} + {\frac{1}{2\cdot3}} = \frac{2}{3}
[/tex]
[tex]
{\frac{1}{1\cdot2}} + {\frac{1}{2\cdot3}} + {\frac{1}{3\cdot4}} = \frac{3}{4}
[/tex]
This leads to the conjecture that,
[tex]
{{\sum_{i = 1}^{n}}{\frac{1}{i(i+1)}}} = {\frac{n}{n+1}}
[/tex]
How would I prove the above by induction?
-PFStudent
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