Difference between revisions of "2006 AMC 12A Problems/Problem 23"
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− | < | + | <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> |
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>. | 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>. |
Revision as of 20:53, 18 December 2020
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
In general, such that
has
terms. Specifically,
To find x, we need only solve the equation
. Algebra yields
.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.