AMC12 2006 12A, Problem 23

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 $x$?

Solution: 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 $x=\fbox{\sqrt{2}-1}$.

Answer: \sqrt{2}-1