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

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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>.
  
−
== Solution ==
+
== Solution 1 ==
−
<math>\dfrac{1}{k^2+k}=\dfrac{1}{k(k+1)}=\dfrac{1}{k}-\dfrac{1}{k+1}</math>. Thus,
+
Using [[partial fraction decomposition]] yields <math>\dfrac{1}{k^2+k}=\dfrac{1}{k(k+1)}=\dfrac{1}{k}-\dfrac{1}{k+1}</math>. Thus,
  
−
<math>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}</math>
+
<math>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}</math>
  
−
Which is
+
Which means that
  
 
<math>\dfrac{n-m}{mn}=\dfrac{1}{29}</math>
 
<math>\dfrac{n-m}{mn}=\dfrac{1}{29}</math>
  
−
Since we need a 29 in the denominator, let <math>n=29t</math>.* Substituting,
+
Since we need a factor of 29 in the denominator, we let <math>n=29t</math>.* Substituting, we get
  
 
<math>29t-m=mt</math>
 
<math>29t-m=mt</math>
 +
 +
so
  
 
<math>\frac{29t}{t+1} = m</math>
 
<math>\frac{29t}{t+1} = m</math>
  
−
Since n is an integer, <math>t+1 = 29</math>, or <math>t=28</math>. It quickly follows that <math>n=29(28)</math> and <math>m=28</math>, so <math>m+n = 30(28) = \fbox{840}</math>.
+
Since <math>m</math> is an integer, <math>t+1 = 29</math>, so <math>t=28</math>. It quickly follows that <math>n=29(28)</math> and <math>m=28</math>, so <math>m+n = 30(28) = \fbox{840}</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>\sum\limits_{i=m}^{n-1} a_i = \sum\limits_{i=1}^{n-1} a_i - \sum\limits_{i=1}^{m-1} a_i = \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, <math>m < n \implies 29 - m = 1</math> and <math>29 + n = 29^2</math>. Hence, <math>m = 28</math>, <math>n = 812</math>, and <math>m+n = \fbox{840}</math>.
 +
 
 +
~ keeper1098
 +
 
 +
== Video Solution by OmegaLearn ==
 +
https://youtu.be/lH-0ul1hwKw?t=134
  
−
*If <math>m=29t</math>, a similar argument to the one above implies <math>m=29(28)</math> and <math>m=28</math>, which implies <math>m>n</math>, which is impossible since <math>n-m>0</math>.
+
~ pi_is_3.14
  
 
== 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}}

Latest revision as of 10:27, 5 October 2024

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

Using partial fraction decomposition yields $\dfrac{1}{k^2+k}=\dfrac{1}{k(k+1)}=\dfrac{1}{k}-\dfrac{1}{k+1}$. 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, $\sum\limits_{i=m}^{n-1} a_i = \sum\limits_{i=1}^{n-1} a_i - \sum\limits_{i=1}^{m-1} a_i = \dfrac{n-1}{n} - \dfrac{m-1}{m} = \dfrac{n-m}{mn} = 1/29$. Cross-multiplying yields $29n - 29m - mn = 0$, and adding $29^2$ to both sides gives $(29-m)(29+n) = 29^2$. Clearly, $m < n \implies 29 - m = 1$ and $29 + n = 29^2$. Hence, $m = 28$, $n = 812$, and $m+n = \fbox{840}$.

~ keeper1098

Video Solution by OmegaLearn

https://youtu.be/lH-0ul1hwKw?t=134

~ pi_is_3.14

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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