Difference between revisions of "2002 AIME II Problems/Problem 6"
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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> | 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. |
== 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}} |
Revision as of 20:41, 12 February 2013
Problem
Find the integer that is closest to .
Solution
You know that .
So if you pull the out of the summation, you get
.
Now that telescopes, leaving you with:
is 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.
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 |