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

Latest revision as of 07:55, 4 November 2022

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 , a similar argument to the one above implies  and , which implies . This is impossible since .

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

@@ Line 1: / Line 1: @@
-{{empty}}
 == Problem ==
 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 ==
-{{solution}}
+<math>\dfrac{1}{k^2+k}=\dfrac{1}{k(k+1)}=\dfrac{1}{k}-\dfrac{1}{k+1}</math>. Thus,
-== See also ==
+<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>
-* [[2002 AIME I Problems/Problem 3| Previous problem]]
+Which means that
+<math>\dfrac{n-m}{mn}=\dfrac{1}{29}</math>
+Since we need a factor of 29 in the denominator, we let <math>n=29t</math>.* Substituting, we get
+<math>29t-m=mt</math>
+so
+<math>\frac{29t}{t+1} = m</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
-* [[2002 AIME I Problems/Problem 5| Next problem]]
+~ pi_is_3.14
-* [[2002 AIME I Problems]]
+== See also ==
+{{AIME box|year=2002|n=I|num-b=3|num-a=5}}
+{{MAA Notice}}

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