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</math> <math>\frac{1}{2}=\frac{1}{3}+\frac{1}{6};\quad \frac{1}{3}=\frac{1}{4}+\frac{1}{12};\ | + | <math>\frac{1}{1}=</math> <math>\frac{1}{2}+</math> <math>\frac{1}{2};</math> <math>\quad</math> <math>\frac{1}{2}=</math> <math>\frac{1}{3}+</math> <math>\frac{1}{6};</math><math>\quad \frac{1}{3}=</math> <math>\frac{1}{4}+\frac{1}{12};\quad...</math> |
State a general law suggested by these examples, and prove it. | State a general law suggested by these examples, and prove it. | ||
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<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 18:56, 27 May 2022
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:
![$\frac{1}{n(n+1)}+\frac{1}{n+1} = \frac{1}{n}$](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 |