Difference between revisions of "2006 AMC 12A Problems/Problem 23"
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\left(\frac{1}{2^{50}}\right)=A^{100}(S)=\left(\frac{(1+x)^{100}}{2^{100}}\right), | \left(\frac{1}{2^{50}}\right)=A^{100}(S)=\left(\frac{(1+x)^{100}}{2^{100}}\right), | ||
</cmath>so <math>(1+x)^{100}=2^{50}</math>, and because <math>x>0</math>, we have <math>x=\boxed{\sqrt{2}-1}</math>. | </cmath>so <math>(1+x)^{100}=2^{50}</math>, and because <math>x>0</math>, we have <math>x=\boxed{\sqrt{2}-1}</math>. | ||
+ | - Alcumus | ||
== See also == | == See also == |
Latest revision as of 21:40, 29 September 2023
Contents
[hide]Problem
Given a finite sequence of real numbers, let be the sequence of real numbers. Define and, for each integer , , define . Suppose , and let . If , then what is ?
Solution 1
In general, such that has terms. Specifically, To find x, we need only solve the equation . Algebra yields .
Solution 2
For every sequence of at least three terms,
Thus for , the coefficients of the terms in the numerator of are the binomial coefficients , and the denominator is . Because for all integers , the coefficients of the terms in the numerators of are for . The definition implies that the denominator of each term in is . For the given sequence, the sole term in is Therefore, so , and because , we have . - Alcumus
See also
2006 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 22 |
Followed by Problem 24 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
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