Difference between revisions of "2006 AMC 12A Problems/Problem 23"
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Given a finite sequence <math>S=(a_1,a_2,\ldots ,a_n)</math> of <math>n</math> real numbers, let <math>A(S)</math> be the sequence | Given a finite sequence <math>S=(a_1,a_2,\ldots ,a_n)</math> of <math>n</math> real numbers, let <math>A(S)</math> be the sequence | ||
+ | <math>\left(\frac{a_1+a_2}{2},\frac{a_2+a_3}{2},\ldots ,\frac{a_{n-1}+a_n}{2}\right)</math> | ||
+ | of <math>n-1</math> real numbers. Define <math>A^1(S)=A(S)</math> and, for each integer <math>m</math>, <math>2\le m\le n-1</math>, define <math>A^m(S)=A(A^{m-1}(S))</math>. Suppose <math>x>0</math>, and let <math>S=(1,x,x^2,\ldots ,x^{100})</math>. If <math>A^{100}(S)=(1/2^{50})</math>, then what is <math>x</math>? | ||
− | <math>(\frac{ | + | <math> \mathrm{(A) \ } 1-\frac{\sqrt{2}}{2}\qquad \mathrm{(B) \ } \sqrt{2}-1\qquad \mathrm{(C) \ } \frac{1}{2}\qquad \mathrm{(D) \ } 2-\sqrt{2}\qquad \mathrm{(E) \ } \frac{\sqrt{2}}{2}</math> |
− | + | == Solution 1 == | |
+ | <cmath>A^1(S)=\left(\frac{1+x}{2},\frac{x+x^2}{2},...,\frac{x^{99}+x^{100}}{2}\right)</cmath> | ||
+ | <cmath>A^2(S)=\left(\frac{1+2x+x^2}{2^2},\frac{x+2x^2+x^3}{2^2},...,\frac{x^{98}+2x^{99}+x^{100}}{2^2}\right)</cmath> | ||
+ | <cmath>\Rightarrow A^2(S)=\left(\frac{(x+1)^2}{2^2},\frac{x(x+1)^2}{2^2},...,\frac{x^{98}(x+1)^2}{2^2}\right)</cmath> | ||
− | <math> | + | In general, <math>A^n(S)=\left(\frac{(x+1)^n}{2^n},\frac{x(x+1)^n}{2^n},...,\frac{x^{100-n}(x+1)^n}{2^n}\right)</math> such that <math>A^n(s)</math> has <math>101-n</math> terms. Specifically, <math>A^{100}(S)=\frac{(x+1)^{100}}{2^{100}}</math> To find x, we need only solve the equation <math>\frac{(x+1)^{100}}{2^{100}}=\frac{1}{2^{50}}</math>. Algebra yields <math>x=\sqrt{2}-1</math>. |
− | == Solution == | + | == Solution 2 == |
− | <math> | + | For every sequence <math>S=\left(a_1,a_2,\dots, |
− | + | a_n\right)</math> of at least three terms, | |
− | |||
− | + | <cmath>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).</cmath>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 <cmath>\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}.</cmath>Therefore, <cmath> | ||
+ | \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>. | ||
+ | - Alcumus | ||
== See also == | == See also == | ||
− | + | ||
− | + | {{AMC12 box|year=2006|ab=A|num-b=22|num-a=24}} | |
− | + | {{MAA Notice}} |
Latest revision as of 22:40, 29 September 2023
Contents
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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