2002 AIME II Problems/Problem 6
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 |
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