Difference between revisions of "2006 AMC 12A Problems/Problem 23"
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== Solution 2 == | == Solution 2 == | ||
− | For every sequence <math>S= | + | For every sequence <math>S=\left(a_1,a_2,\dots, |
− | a_n | + | a_n\right)</math> of at least three terms, |
− | \[ | + | <math>\[ |
− | A^2(S)= | + | A^2(S)=\left(\frac{a_1+2a_2+a_3}{4},\frac{a_2+2a_3+a_4}{4},\dots,\frac{a_{n-2}+2a_{n-1}+a_n}{4}\right).\]</math>Thus for <math>m = 1\text{ and }2</math>, the coefficients of the terms in the numerator of <math>A^m(S)</math> are the binomial coefficients <math>\binom{m}{0},\binom{m}{1},\dots,\binom{m}{m}</math>, and the denominator is <math>2^m</math>. Because <math>\binom{m}{r}+\binom{m}{r+1}=\binom{m+1}{r+1}</math> for all integers <math>r\geq 0</math>, the coefficients of the terms in the numerators of <math>A^{m+1}(S)</math> are <math>\binom{m+1}{0},\binom{m+1}{1},\ldots,\binom{m+1}{m+1}</math> for <math>2\leq |
m\leq n-2</math>. The definition implies that the denominator of each term in <math>A^{m+1}(S)</math> is <math>2^{m+1}</math>. For the given sequence, the sole term in <math>A^{100}(S)</math> is\[ \frac{1}{2^{100}} \sum_{m=0}^{100} \binom{100}{m}a_{m+1} = | m\leq n-2</math>. The definition implies that the denominator of each term in <math>A^{m+1}(S)</math> is <math>2^{m+1}</math>. For the given sequence, the sole term in <math>A^{100}(S)</math> is\[ \frac{1}{2^{100}} \sum_{m=0}^{100} \binom{100}{m}a_{m+1} = | ||
\frac{1}{2^{100}} \sum_{m=0}^{100} \binom{100}{m}x^m = | \frac{1}{2^{100}} \sum_{m=0}^{100} \binom{100}{m}x^m = | ||
\frac{1}{2^{100}}(x+1)^{100}.\]Therefore\[ | \frac{1}{2^{100}}(x+1)^{100}.\]Therefore\[ | ||
− | + | \left(\frac{1}{2^{50}}\right)=A^{100}(S)=\left(\frac{(1+x)^{100}}{2^{100}}\right), | |
\]so <math>(1+x)^{100}=2^{50}</math>, and because <math>x>0</math>, we have <math>x=\boxed{\sqrt{2}-1}</math>. | \]so <math>(1+x)^{100}=2^{50}</math>, and because <math>x>0</math>, we have <math>x=\boxed{\sqrt{2}-1}</math>. | ||
- Alcumus | - Alcumus |
Revision as of 20:41, 27 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,
$\[ A^2(S)=\left(\frac{a_1+2a_2+a_3}{4},\frac{a_2+2a_3+a_4}{4},\dots,\frac{a_{n-2}+2a_{n-1}+a_n}{4}\right).\]$ (Error compiling LaTeX. Unknown error_msg)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\[ \frac{1}{2^{100}} \sum_{m=0}^{100} \binom{100}{m}a_{m+1} = \frac{1}{2^{100}} \sum_{m=0}^{100} \binom{100}{m}x^m = \frac{1}{2^{100}}(x+1)^{100}.\]Therefore\[ \left(\frac{1}{2^{50}}\right)=A^{100}(S)=\left(\frac{(1+x)^{100}}{2^{100}}\right), \]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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