2002 AIME II Problems/Problem 6
Find the integer that is closest to .
We know that . We can use the process of fractional decomposition to split this into two fractions: 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
And so, .
This telescopes into:
The small fractional terms are not enough to bring lower than so the answer is
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
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