Question

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

$1 - \dfrac{\sqrt{2}}{2}$

$\sqrt{2} - 1$

$\dfrac{1}{2}$

$2 - \sqrt{2}$

$\dfrac{\sqrt{2}}{2}$

Answer: B

Concepts:binomial theorem recursion

Difficulty rating: 2400

Solution:

Each application of $A$ averages adjacent terms, so after $100$ steps the single remaining term is $\begin{gathered} \frac{1}{2^{100}} \sum_{m=0}^{100} \binom{100}{m} x^m \\ = \frac{(1 + x)^{100}}{2^{100}}. \end{gathered}$

Setting this equal to $\dfrac{1}{2^{50}}$ gives $(1 + x)^{100} = 2^{50},$ so $1 + x = 2^{1/2} = \sqrt{2}.$ Since $x \gt 0,$ we get $x = \sqrt{2} - 1.$

Thus, the correct answer is B.

2006 AMC 12A Problem 23

Answer: B

Difficulty rating: 2400

Solution:

Problem 23 in Other Years