Difference between revisions of "1973 Canadian MO Problems/Problem 7"
Pi3point14 (talk | contribs) m |
Hastapasta (talk | contribs) |
||
Line 9: | Line 9: | ||
<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> | <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> | ||
− | ==Solution== | + | |
+ | == Solution 1 (Easiest) == | ||
+ | |||
+ | <math>\text{i}</math>: We see that <math>\frac{1}{n(n+1)}+\frac{1}{n+1} = \frac{1}{n}</math>. By simply evaluating the left hand side using common denominators, we see that both sides are equal. So therefore we have proved the law. | ||
+ | |||
+ | <math>\text{ii}</math>: Using partial fraction decomposition, let's change the original expression to: | ||
+ | |||
+ | <math>\frac{1}{i}-\frac{1}{i+1}+\frac{1}{i+1}-\frac{1}{i+2}+...+\frac{1}{j}-\frac{1}{j+1}=\frac{1}{i}-\frac{j+1}=\frac{1}{n}</math>. | ||
+ | |||
+ | Therefore, <math>\frac{1}{i}=\frac{1}{n}+\frac{1}{j+1}</math>. | ||
+ | |||
+ | Let <math>i=(j+1)(j+2), n=j+2</math>, then we have satisfied the previous identity in part <math>\text{i}</math>. Therefore, we have solved the problem. | ||
+ | |||
+ | ~hastapasta | ||
+ | |||
+ | ==Solution 2 == | ||
<math>\text{(i)}</math> We see that: <center><math>\frac{1}{n(n+1)}+\frac{1}{n+1} = \frac{1}{n}</math></center> | <math>\text{(i)}</math> We see that: <center><math>\frac{1}{n(n+1)}+\frac{1}{n+1} = \frac{1}{n}</math></center> |
Latest revision as of 17:56, 27 May 2022
Contents
[hide]Problem
Observe that 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 1 (Easiest)
: We see that . By simply evaluating the left hand side using common denominators, we see that both sides are equal. So therefore we have proved the law.
: Using partial fraction decomposition, let's change the original expression to:
.
Therefore, .
Let , then we have satisfied the previous identity in part . Therefore, we have solved the problem.
~hastapasta
Solution 2
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 |