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

Revision as of 23:53, 9 November 2010

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

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

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

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

$29t-m=mt$

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

Since n is an integer, $t+1 = 29$ , or $t=28$ . 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 $m=28$ , which implies $m>n$ , which is impossible since $n-m>0$ .

@@ Line 11: / Line 11: @@
 <math>\dfrac{n-m}{mn}=\dfrac{1}{29}</math>
-Since we need a 29 in the denominator, let <math>m=29t</math>.* Substituting,
+Since we need a 29 in the denominator, let <math>n=29t</math>.* Substituting,
 <math>29t-m=mt</math>

2002 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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Difference between revisions of "2002 AIME I Problems/Problem 4"

Revision as of 23:53, 9 November 2010

Problem

Solution

See also