Difference between revisions of "2002 AIME II Problems/Problem 6"
(→Solution) |
m (Breaking into sections) |
||
(12 intermediate revisions by 11 users not shown) | |||
Line 2: | Line 2: | ||
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>. | ||
− | == Solution == | + | == Solution 1 == |
− | + | We know that <math>\frac{4}{n^2 - 4} = \frac{1}{n-2} - \frac{1}{n + 2}</math>. | |
− | So if | + | So if we pull the <math>\frac{1}{4}</math> out of the summation, you get |
<math>250\sum_{n=3}^{10,000} (\frac{1}{n-2} - \frac{1}{n + 2})</math>. | <math>250\sum_{n=3}^{10,000} (\frac{1}{n-2} - \frac{1}{n + 2})</math>. | ||
Line 11: | Line 11: | ||
Now that telescopes, leaving you with: | Now that telescopes, leaving you with: | ||
− | <math>250 (1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} - \frac{1}{ | + | <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> |
− | + | 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> | |
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: | 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: | ||
− | 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 | + | 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. |
− | Solving for A and B gives <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> |
− | Then we have <math>\frac{1}{(n+2)(n-2)} = \frac{ \frac{1}{4} }{(n | + | 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 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}} |
Latest revision as of 14:55, 9 August 2020
Contents
Problem
Find the integer that is closest to .
Solution 1
We know that .
So if we pull the out of the summation, you get
.
Now that telescopes, leaving you with:
The small fractional terms are not enough to bring lower than so the answer is
If you didn't know , here's how you can find it out:
We know . We can use the process of fractional decomposition to split this into two fractions thus: for some A and B.
Solving for A and B gives or . Since there is no n term on the left hand side, and by inspection . Solving yields
Then we have and we can continue as before.
Solution 2
Using the fact that or by partial fraction decomposition, we both obtained . The denominators of the positive terms are , while the negative ones are . Hence we are left with . We can simply ignore the last terms, and we get it is approximately . Computing which is about and moving the decimal point three times, we get that the answer is
See also
2002 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.