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Difference between revisions of "2002 AIME I Problems/Problem 4"

m (Changed the wording a tiny bit again to make it clearer.)
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Consider the sequence defined by <math>a_k =\dfrac{1}{k^2+k}</math> for <math>k\geq 1</math>. Given that <math>a_m+a_{m+1}+\cdots+a_{n-1}=\dfrac{1}{29}</math>, for positive integers <math>m</math> and <math>n</math> with <math>m<n</math>, find <math>m+n</math>.
 
Consider the sequence defined by <math>a_k =\dfrac{1}{k^2+k}</math> for <math>k\geq 1</math>. Given that <math>a_m+a_{m+1}+\cdots+a_{n-1}=\dfrac{1}{29}</math>, for positive integers <math>m</math> and <math>n</math> with <math>m<n</math>, find <math>m+n</math>.
  
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== Solution ==
+
== Solution 1 ==
 
<math>\dfrac{1}{k^2+k}=\dfrac{1}{k(k+1)}=\dfrac{1}{k}-\dfrac{1}{k+1}</math>. Thus,
 
<math>\dfrac{1}{k^2+k}=\dfrac{1}{k(k+1)}=\dfrac{1}{k}-\dfrac{1}{k+1}</math>. Thus,
  
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  *If <math>m=29t</math>, a similar argument to the one above implies <math>m=29(28)</math> and <math>n=28</math>, which implies <math>m>n</math>. This is impossible since <math>n-m>0</math>.
 
  *If <math>m=29t</math>, a similar argument to the one above implies <math>m=29(28)</math> and <math>n=28</math>, which implies <math>m>n</math>. This is impossible since <math>n-m>0</math>.
 +
 +
 +
== Solution 2 ==
 +
Note that <math>a_1 + a_2 + \cdots + a_i = \dfrac{i}{i+1}</math>. This can be proven by induction. Thus, <math></math>\sum_{i=m}^{\n-1} a_i<math> = </math>\sum_{i=1}^{\n-1} a_i<math> - </math>\sum_{i=1}^{\m-1} a_i<math> = \dfrac{n-1}{n} - \dfrac{m-1}{m} = \dfrac{n-m}{mn} = 1/29.</math> Cross-multiplying yields <math>29n - 29m - mn = 0</math>, and adding <math>29^2</math> to both sides gives <math>(29-m)(29+n) = 29^2</math>. Clearly, since <math>m < n</math>, <math>29-m = 1</math> and <math>29+n</math> = <math>29^2</math>. Hence, <math>m = 28</math>, <math>n = 812</math>, and <math>m+n = \fbox{840}</math>.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2002|n=I|num-b=3|num-a=5}}
 
{{AIME box|year=2002|n=I|num-b=3|num-a=5}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 17:59, 4 May 2020

Problem

Consider the sequence defined by $a_k =\dfrac{1}{k^2+k}$ for $k\geq 1$. Given that $a_m+a_{m+1}+\cdots+a_{n-1}=\dfrac{1}{29}$, for positive integers $m$ and $n$ with $m<n$, find $m+n$.

Solution 1

$\dfrac{1}{k^2+k}=\dfrac{1}{k(k+1)}=\dfrac{1}{k}-\dfrac{1}{k+1}$. Thus,

$a_m+a_{m+1}+\cdots +a_{n-1}=\dfrac{1}{m}-\dfrac{1}{m+1}+\dfrac{1}{m+1}-\dfrac{1}{m+2}+\cdots +\dfrac{1}{n-1}-\dfrac{1}{n}=\dfrac{1}{m}-\dfrac{1}{n}$

Which means that

$\dfrac{n-m}{mn}=\dfrac{1}{29}$

Since we need a factor of 29 in the denominator, we let $n=29t$.* Substituting, we get

$29t-m=mt$

so

$\frac{29t}{t+1} = m$

Since $m$ is an integer, $t+1 = 29$, so $t=28$. It quickly follows that $n=29(28)$ and $m=28$, so $m+n = 30(28) = \fbox{840}$. 

*If $m=29t$, a similar argument to the one above implies $m=29(28)$ and $n=28$, which implies $m>n$. This is impossible since $n-m>0$.


Solution 2

Note that $a_1 + a_2 + \cdots + a_i = \dfrac{i}{i+1}$. This can be proven by induction. Thus, $$ (Error compiling LaTeX. Unknown error_msg)\sum_{i=m}^{\n-1} a_i$=$\sum_{i=1}^{\n-1} a_i$-$\sum_{i=1}^{\m-1} a_i$= \dfrac{n-1}{n} - \dfrac{m-1}{m} = \dfrac{n-m}{mn} = 1/29.$ Cross-multiplying yields $29n - 29m - mn = 0$, and adding $29^2$ to both sides gives $(29-m)(29+n) = 29^2$. Clearly, since $m < n$, $29-m = 1$ and $29+n$ = $29^2$. Hence, $m = 28$, $n = 812$, and $m+n = \fbox{840}$.

See also

2002 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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