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Difference between revisions of "2002 AIME II Problems/Problem 6"

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Find the integer that is closest to <math>1000\sum_{n=3}^{10000}\frac1{n^2-4}</math>.
 
Find the integer that is closest to <math>1000\sum_{n=3}^{10000}\frac1{n^2-4}</math>.
  
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== Solution ==
+
== Solution 1 ==
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We know that <math>\frac{4}{n^2 - 4} = \frac{1}{n-2} - \frac{1}{n + 2}</math>.
+
We know that <math>\frac{1}{n^2 - 4} = \frac{1}{(n+2)(n-2)}</math>. We can use the process of fractional decomposition to split this into two fractions: <math>\frac{1}{(n+2)(n-2)} = \frac{A}{(n+2)} + \frac{B}{(n-2)}</math> for some A and B.  
  
−
So if we pull the <math>\frac{1}{4}</math> out of the summation, you get
+
Solving for A and B gives <math>1 = (n-2)A + (n+2)B</math> or <math>1 = n(A+B)+ 2(B-A)</math>. Since there is no n term on the left hand side, <math> A+B=0</math> and by inspection <math>1 = 2(B-A)</math>. Solving yields <math> A=\frac{1}{4},  B=\frac{-1}{4}</math>
−
 
 
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<math>250\sum_{n=3}^{10,000} (\frac{1}{n-2} - \frac{1}{n + 2})</math>.
 
−
 
 
−
Now that telescopes, leaving you with:
 
−
 
 
−
<math>250 (1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} - \frac{1}{9999} - \frac{1}{10000} - \frac{1}{10001} - \frac{1}{10002}) = 250 + 125 + 83.3 + 62.5 - 250 (- \frac{1}{9999} - \frac{1}{10000} - \frac{1}{10001} - \frac{1}{10002})</math>
 
  
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<math>250(-\frac{1}{9997} - \frac{1}{9998} - \frac{1}{9999} - \frac{1}{10000})</math> is not enough to bring <math>520.8</math> lower than <math>520.5</math>  so the answer is <math>\fbox{521}</math>
+
Therefore, <math>\frac{1}{n^2-4} = \frac{1}{(n+2)(n-2)} = \frac{ \frac{1}{4} }{(n-2)} + \frac{ \frac{-1}{4} }{(n+2)} = \frac{1}{4} \left( \frac{1}{n-2} - \frac{1}{n+2} \right)</math>.
  
 +
And so, <math>1000\sum_{n=3}^{10,000} \frac{1}{n^2-4} = 1000\sum_{n=3}^{10,000} \frac{1}{4} \left( \frac{1}{n-2} - \frac{1}{n+2} \right) = 250\sum_{n=3}^{10,000} (\frac{1}{n-2} - \frac{1}{n + 2})</math>.
  
−
If you didn't know <math>\frac{4}{n^2 - 4} = \frac{1}{n-2} - \frac{1}{n + 2}</math>, here's how you can find it out:
+
This telescopes into:
  
−
We know <math>\frac{1}{n^2 - 4} = \frac{1}{(n+2)(n-2)}</math>. We can use the process of fractional decomposition to split this into two fractions thus: <math>\frac{1}{(n+2)(n-2)} = \frac{A}{(n+2)} + \frac{B}{(n+2)}</math> for some A and B.
+
<math>250 (1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} - \frac{1}{9999} - \frac{1}{10000} - \frac{1}{10001} - \frac{1}{10002}) = 250 + 125 + 83.3 + 62.5 - 250 (\frac{1}{9999} + \frac{1}{10000} + \frac{1}{10001} + \frac{1}{10002})</math>
  
−
Solving for A and B gives <math>1 = (n-2)A + (n+2)B</math> or <math>1 = n(A+B)+ 2(B-A)</math>. Since there is no n term on the left hand side, <math> A+B=0</math> and by inspection <math>1 = 2(B-A)</math>. Solving yields <math> A=\frac{1}{4},  B=\frac{-1}{4}</math>
+
The small fractional terms are not enough to bring <math>520.8</math> lower than <math>520.5,</math> so the answer is <math>\fbox{521}</math>
  
−
Then we have <math>\frac{1}{(n+2)(n-2)} = \frac{ \frac{1}{4} }{(n-2)} + \frac{ \frac{-1}{4} }{(n+2)}</math> and we can continue as before.
+
== Solution 2 ==
 +
Using the fact that <math>\frac{1}{n(n+k)} = \frac{1}{k} ( \frac{1}{n}-\frac{1}{n+k} )</math> or by partial fraction decomposition, we both obtained <math>\frac{1}{x^2-4} = \frac{1}{4}(\frac{1}{x-2}-\frac{1}{x+2})</math>. The denominators of the positive terms are <math>1,2,..,9998</math>, while the negative ones are <math>5,6,...,10002</math>. Hence we are left with <math>1000 \cdot \frac{1}{4} (1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} - \frac{1}{9999} - \frac{1}{10000} - \frac{1}{10001} - \frac{1}{10002})</math>. We can simply ignore the last <math>4</math> terms, and we get it is approximately <math>1000*\frac{25}{48}</math>. Computing <math>\frac{25}{48}</math> which is about <math>0.5208..</math> and moving the decimal point three times, we get that the answer is <math>521</math>
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2002|n=II|num-b=5|num-a=7}}
 
{{AIME box|year=2002|n=II|num-b=5|num-a=7}}
 +
 +
[[Category: Intermediate Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}
−
Although the answer to Problem 6 doesn't change, the value of the telescoping sum is incorrect as given.  Instead of
 
−
<cmath>250 \left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{9997}-\frac{1}{9998}-\frac{1}{9999}-\frac{1}{10000} \right), </cmath>
 
−
the correct sum is
 
−
<cmath>250 \left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{9999}-\frac{1}{10000}-\frac{1}{10001}-\frac{1}{10002} \right). </cmath>
 

Latest revision as of 04:56, 23 August 2022

Problem

Find the integer that is closest to $1000\sum_{n=3}^{10000}\frac1{n^2-4}$.

Solution 1

We know that $\frac{1}{n^2 - 4} = \frac{1}{(n+2)(n-2)}$. We can use the process of fractional decomposition to split this into two fractions: $\frac{1}{(n+2)(n-2)} = \frac{A}{(n+2)} + \frac{B}{(n-2)}$ for some A and B.

Solving for A and B gives $1 = (n-2)A + (n+2)B$ or $1 = n(A+B)+ 2(B-A)$. Since there is no n term on the left hand side, $A+B=0$ and by inspection $1 = 2(B-A)$. Solving yields $A=\frac{1}{4}, B=\frac{-1}{4}$

Therefore, $\frac{1}{n^2-4} = \frac{1}{(n+2)(n-2)} = \frac{ \frac{1}{4} }{(n-2)} + \frac{ \frac{-1}{4} }{(n+2)} = \frac{1}{4} \left( \frac{1}{n-2} - \frac{1}{n+2} \right)$.

And so, $1000\sum_{n=3}^{10,000} \frac{1}{n^2-4} = 1000\sum_{n=3}^{10,000} \frac{1}{4} \left( \frac{1}{n-2} - \frac{1}{n+2} \right) = 250\sum_{n=3}^{10,000} (\frac{1}{n-2} - \frac{1}{n + 2})$.

This telescopes into:

$250 (1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} - \frac{1}{9999} - \frac{1}{10000} - \frac{1}{10001} - \frac{1}{10002}) = 250 + 125 + 83.3 + 62.5 - 250 (\frac{1}{9999} + \frac{1}{10000} + \frac{1}{10001} + \frac{1}{10002})$

The small fractional terms are not enough to bring $520.8$ lower than $520.5,$ so the answer is $\fbox{521}$

Solution 2

Using the fact that $\frac{1}{n(n+k)} = \frac{1}{k} ( \frac{1}{n}-\frac{1}{n+k} )$ or by partial fraction decomposition, we both obtained $\frac{1}{x^2-4} = \frac{1}{4}(\frac{1}{x-2}-\frac{1}{x+2})$. The denominators of the positive terms are $1,2,..,9998$, while the negative ones are $5,6,...,10002$. Hence we are left with $1000 \cdot \frac{1}{4} (1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} - \frac{1}{9999} - \frac{1}{10000} - \frac{1}{10001} - \frac{1}{10002})$. We can simply ignore the last $4$ terms, and we get it is approximately $1000*\frac{25}{48}$. Computing $\frac{25}{48}$ which is about $0.5208..$ and moving the decimal point three times, we get that the answer is $521$

See also

2002 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
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All AIME Problems and Solutions

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