# Difference between revisions of "2006 AMC 12A Problems/Problem 23"

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== Solution == | == Solution == | ||

− | {{ | + | <math>A^1(S)=\left(\frac{1+x}{2},\frac{x+x^2}{2},...,\frac{x^{99}+x^{100}}{2}\right)</math> |

+ | <math>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)</math> | ||

+ | <math>\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)</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>. | ||

== See also == | == See also == |

## Revision as of 21:36, 14 January 2007

## 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 .