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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

Problem

Given a finite sequence $S=(a_1,a_2,\ldots ,a_n)$ of $n$ real numbers, let $A(S)$ be the sequence $\left(\frac{a_1+a_2}{2},\frac{a_2+a_3}{2},\ldots ,\frac{a_{n-1}+a_n}{2}\right)$ of $n-1$ real numbers. Define $A^1(S)=A(S)$ and, for each integer $m$, $2\le m\le n-1$, define $A^m(S)=A(A^{m-1}(S))$. Suppose $x>0$, and let $S=(1,x,x^2,\ldots ,x^{100})$. If $A^{100}(S)=(1/2^{50})$, then what is $x$?

$\mathrm{(A) \ } 1-\frac{\sqrt{2}}{2}\qquad \mathrm{(B) \ } \sqrt{2}-1\qquad \mathrm{(C) \ } \frac{1}{2}\qquad \mathrm{(D) \ } 2-\sqrt{2}\qquad \mathrm{(E) \ } \frac{\sqrt{2}}{2}$

Solution 1

\[A^1(S)=\left(\frac{1+x}{2},\frac{x+x^2}{2},...,\frac{x^{99}+x^{100}}{2}\right)\] \[A^2(S)=\left(\frac{1+2x+x^2}{2^2},\frac{x+2x^2+x^3}{2^2},...,\frac{x^{98}+2x^{99}+x^{100}}{2^2}\right)\] \[\Rightarrow A^2(S)=\left(\frac{(x+1)^2}{2^2},\frac{x(x+1)^2}{2^2},...,\frac{x^{98}(x+1)^2}{2^2}\right)\]

In general, $A^n(S)=\left(\frac{(x+1)^n}{2^n},\frac{x(x+1)^n}{2^n},...,\frac{x^{100-n}(x+1)^n}{2^n}\right)$ such that $A^n(s)$ has $101-n$ terms. Specifically, $A^{100}(S)=\frac{(x+1)^{100}}{2^{100}}$ To find x, we need only solve the equation $\frac{(x+1)^{100}}{2^{100}}=\frac{1}{2^{50}}$. Algebra yields $x=\sqrt{2}-1$.

Solution 2

For every sequence $S=\left(a_1,a_2,\dots, a_n\right)$ of at least three terms,

\[A^2(S)=\left(\frac{a_1+2a_2+a_3}{4},\frac{a_2+2a_3+a_4}{4},\dots,\frac{a_{n-2}+2a_{n-1}+a_n}{4}\right).\]Thus for $m = 1\text{ and }2$, the coefficients of the terms in the numerator of $A^m(S)$ are the binomial coefficients $\binom{m}{0},\binom{m}{1},\dots,\binom{m}{m}$, and the denominator is $2^m$. Because $\binom{m}{r}+\binom{m}{r+1}=\binom{m+1}{r+1}$ for all integers $r\geq 0$, the coefficients of the terms in the numerators of $A^{m+1}(S)$ are $\binom{m+1}{0},\binom{m+1}{1},\ldots,\binom{m+1}{m+1}$ for $2\leq m\leq n-2$. The definition implies that the denominator of each term in $A^{m+1}(S)$ is $2^{m+1}$. For the given sequence, the sole term in $A^{100}(S)$ 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 $(1+x)^{100}=2^{50}$, and because $x>0$, we have $x=\boxed{\sqrt{2}-1}$. - Alcumus

See also

2006 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
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All AMC 12 Problems and Solutions

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