Difference between revisions of "2000 AIME II Problems/Problem 7"
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− | == Solution | + | == Solution 2 == |
Let <math>f(x) = (1+x)^{19}.</math> Applying the binomial theorem gives us <math>f(x) = \binom{19}{19} x^{19} + \binom{19}{18} x^{18} + \binom{19}{17} x^{17}+ \cdots + \binom{19}{0}.</math> Since <math>\frac 1{2!17!}+\frac 1{3!16!}+\dots+\frac 1{8!11!}+\frac 1{9!10!} = \frac{\frac{f(1)}{2} - \binom{19}{19} - \binom{19}{18}}{19!},</math> <math>N = \frac{2^{18}-20}{19}.</math> After some fairly easy bashing, we get <math>\boxed{137}</math> as the answer. | Let <math>f(x) = (1+x)^{19}.</math> Applying the binomial theorem gives us <math>f(x) = \binom{19}{19} x^{19} + \binom{19}{18} x^{18} + \binom{19}{17} x^{17}+ \cdots + \binom{19}{0}.</math> Since <math>\frac 1{2!17!}+\frac 1{3!16!}+\dots+\frac 1{8!11!}+\frac 1{9!10!} = \frac{\frac{f(1)}{2} - \binom{19}{19} - \binom{19}{18}}{19!},</math> <math>N = \frac{2^{18}-20}{19}.</math> After some fairly easy bashing, we get <math>\boxed{137}</math> as the answer. | ||
− | + | ~peelybonehead | |
== See also == | == See also == |
Revision as of 17:08, 11 August 2022
Contents
[hide]Problem
Given that
find the greatest integer that is less than .
Solution
Multiplying both sides by yields:
Recall the Combinatorial Identity . Since , it follows that .
Thus, .
So, and .
Solution 2
Let Applying the binomial theorem gives us Since After some fairly easy bashing, we get as the answer.
~peelybonehead
See also
2000 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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