Difference between revisions of "1973 Canadian MO Problems"

(Problem 7)
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==Problem 7==
 
==Problem 7==
  
 +
Observe that
 +
<math>\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...</math>
 +
State a general law suggested by these examples, and prove it.
  
 +
Prove that for any integer <math>n</math> greater than <math>1</math> there exist positive integers <math>i</math> and <math>j</math> such that
 +
<math>\frac{1}{n}= \frac{1}{i(i+1)}+\frac{1}{(i+1)(i+2)}+\frac{1}{(i+2)(i+3)}+\cdots+\frac{1}{j(j+1)}. </math>
  
 
[[1973 Canadian MO Problems/Problem 7 | Solution]]
 
[[1973 Canadian MO Problems/Problem 7 | Solution]]

Revision as of 20:55, 16 December 2011

Problem 1

$\text{(i)}$ Solve the simultaneous inequalities, $x<\frac{1}{4x}$ and $x<0$; i.e. find a single inequality equivalent to the two simultaneous inequalities.

$\text{(ii)}$ What is the greatest integer that satisfies both inequalities $4x+13 < 0$ and $x^{2}+3x > 16$.

$\text{(iii)}$ Give a rational number between $11/24$ and $6/13$.

$\text{(iv)}$ Express $100000$ as a product of two integers neither of which is an integral multiple of $10$.

$\text{(v)}$ Without the use of logarithm tables evaluate $\frac{1}{\log_{2}36}+\frac{1}{\log_{3}36}$.


Solution

Problem 2

Solution

Problem 3

Solution

Problem 4

Solution

Problem 5

Solution

Problem 6

Solution

Problem 7

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 $n$ greater than $1$ there exist positive integers $i$ and $j$ such that $\frac{1}{n}= \frac{1}{i(i+1)}+\frac{1}{(i+1)(i+2)}+\frac{1}{(i+2)(i+3)}+\cdots+\frac{1}{j(j+1)}.$

Solution

Resources

1973 Canadian MO