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

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~ keeper1098
 
~ keeper1098
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== Solution 3 ==
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To solve this problem, I need to find two positive integers <math>m</math> and <math>n</math> where <math>m < n</math> and the sum of sequence terms equals <math>\frac{1}{29}</math>.
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 +
First, let me simplify <math>a_k = \frac{1}{k^2 + k}</math> using partial fractions.
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<math>a_k = \frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}</math>
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 +
Express the sum <math>a_m + a_{m+1} + \cdots + a_{n-1}</math> using this simplification.
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<math>a_m + a_{m+1} + \cdots + a_{n-1} = \sum_{k=m}^{n-1} \left(\frac{1}{k} - \frac{1}{k+1}\right)</math>
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 +
This is a telescoping series where intermediate terms cancel:
 +
<math>a_m + a_{m+1} + \cdots + a_{n-1} = \frac{1}{m} - \frac{1}{n}</math>
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 +
Use the given condition that this sum equals <math>\frac{1}{29}</math>.
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<math>\frac{1}{m} - \frac{1}{n} = \frac{1}{29}</math>
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Multiplying both sides by <math>mn</math>:
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<math>n - m = \frac{mn}{29}</math>
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Rearranging:
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<math>29(n - m) = mn</math>
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<math>29n - 29m = mn</math>
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<math>29n - mn = 29m</math>
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<math>n(29 - m) = 29m</math>
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Solve for <math>n</math> in terms of <math>m</math>.
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<math>n = \frac{29m}{29-m}</math>
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Since <math>n</math> must be a positive integer, <math>29-m</math> must divide <math>29m</math> evenly.
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Since <math>29</math> is prime, for <math>29-m</math> to divide <math>29m</math> (when <math>m < 29</math>), we need <math>29-m</math> to divide <math>m</math>.
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This means <math>m = k(29-m)</math> for some positive integer <math>k</math>.
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<math>m = k(29-m)</math>
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<math>m = 29k - km</math>
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<math>m(1+k) = 29k</math>
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<math>m = \frac{29k}{1+k}</math>
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For <math>m</math> to be an integer, <math>1+k</math> must divide <math>29k</math>.
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When <math>k = 28</math>, we get <math>m = \frac{29(28)}{29} = 28</math>
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Calculate <math>n</math> using our value of <math>m</math>.
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<math>n = \frac{29(28)}{29-28} = \frac{812}{1} = 812</math>
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Therefore, <math>m + n = 28 + 812 = 840</math>
  
 
== Video Solution by OmegaLearn ==
 
== Video Solution by OmegaLearn ==

Revision as of 22:25, 6 March 2025

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

Solution 3

To solve this problem, I need to find two positive integers $m$ and $n$ where $m < n$ and the sum of sequence terms equals $\frac{1}{29}$.

First, let me simplify $a_k = \frac{1}{k^2 + k}$ using partial fractions. $a_k = \frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}$

Express the sum $a_m + a_{m+1} + \cdots + a_{n-1}$ using this simplification. $a_m + a_{m+1} + \cdots + a_{n-1} = \sum_{k=m}^{n-1} \left(\frac{1}{k} - \frac{1}{k+1}\right)$

This is a telescoping series where intermediate terms cancel: $a_m + a_{m+1} + \cdots + a_{n-1} = \frac{1}{m} - \frac{1}{n}$

Use the given condition that this sum equals $\frac{1}{29}$. $\frac{1}{m} - \frac{1}{n} = \frac{1}{29}$

Multiplying both sides by $mn$: $n - m = \frac{mn}{29}$

Rearranging: $29(n - m) = mn$ $29n - 29m = mn$ $29n - mn = 29m$ $n(29 - m) = 29m$

Solve for $n$ in terms of $m$. $n = \frac{29m}{29-m}$

Since $n$ must be a positive integer, $29-m$ must divide $29m$ evenly. Since $29$ is prime, for $29-m$ to divide $29m$ (when $m < 29$), we need $29-m$ to divide $m$.

This means $m = k(29-m)$ for some positive integer $k$. $m = k(29-m)$ $m = 29k - km$ $m(1+k) = 29k$ $m = \frac{29k}{1+k}$

For $m$ to be an integer, $1+k$ must divide $29k$. When $k = 28$, we get $m = \frac{29(28)}{29} = 28$

Calculate $n$ using our value of $m$. $n = \frac{29(28)}{29-28} = \frac{812}{1} = 812$

Therefore, $m + n = 28 + 812 = 840$

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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