Difference between revisions of "1973 Canadian MO Problems/Problem 7"
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Revision as of 21:44, 16 December 2011
Problem
Observe that
$\frac{1}{1}= \frac{1}{2}+\frac{1}{2};\quad \frac{1}{2}=\frac{1}{3}+\frac{1}{6};\quad \frac{1}{3}=\frac{1}{4}+\frac{1}{12};\qu...$ (Error compiling LaTeX. Unknown error_msg)
State a general law suggested by these examples, and prove it.
Prove that for any integer
greater than
there exist positive integers
and
such that
Solution
We see that:

We prove this by induction. Let
Base case:
Therefore,
is true.
Now, assume that
is true for some
. Then:

Thus, by induction, the formula holds for all
Incomplete
See also
1973 Canadian MO (Problems) | ||
Preceded by Problem 6 |
1 • 2 • 3 • 4 • 5 | Followed by Problem 1 |