Difference between revisions of "1973 Canadian MO Problems/Problem 7"
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<math>\text{(i)}</math> Observe that | <math>\text{(i)}</math> Observe that | ||
− | <math>\frac{1}{1}=</math> <math>\frac{1}{2}+</math> <math>\frac{1}{2};</math><math>\quad \frac{1}{2}=\frac{1}{3}+\frac{1}{6};\quad \frac{1}{3}=\frac{1}{4}+\frac{1}{12};\qu...</math> | + | <math>\frac{1}{1}=</math> <math>\frac{1}{2}+</math> <math>\frac{1}{2};</math> <math>\quad</math> <math>\frac{1}{2}=\frac{1}{3}+\frac{1}{6};\quad \frac{1}{3}=\frac{1}{4}+\frac{1}{12};\qu...</math> |
State a general law suggested by these examples, and prove it. | State a general law suggested by these examples, and prove it. | ||
Revision as of 20:05, 4 December 2015
Problem
Observe that
$\frac{1}{2}=\frac{1}{3}+\frac{1}{6};\quad \frac{1}{3}=\frac{1}{4}+\frac{1}{12};\qu...$ (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:
![$\frac{1}{n(n+1)}+\frac{1}{n+1} = \frac{1}{n}$](http://latex.artofproblemsolving.com/5/4/e/54e83a6d9fd85bef18b3dd6dbc7837e2c638ad16.png)
We prove this by induction. Let
Base case:
Therefore,
is true.
Now, assume that
is true for some
. Then:
![$\begin{matrix} \text{LHS of } P(k+1) &=& \frac{1}{(k+1)((k+1)+1)}+\frac{1}{(k+1)+1}\\ &=& \frac{1}{(k+1)(k+2)}+\frac{1}{k+2}\\ &=& \frac{1}{(k+1)(k+2)}+\frac{k+1}{(k+2)(k+1)}\\ &=& \frac{k+2}{(k+1)(k+2)}\\ &=& \frac{1}{k+1} &=& \text{RHS of } P(k+1)\\\end{matrix}$](http://latex.artofproblemsolving.com/4/7/f/47f63a4039d3a52590a5e3b0a545cc5628496abe.png)
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 |