2006 AMC 12A Problem 23

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) = (\frac{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}

Applying

A

repeatedly introduces binomial coefficients;

A^{100}(S)

has a single term

That term is

\dfrac{(1 + x)^{100}}{2^{100}}

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^{\frac{1}{2}} = \sqrt{2}.

Since

x \gt 0,

we get

x = \sqrt{2} - 1.

Thus, the correct answer is B.

Problem 23 in other years