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2006 AMC 12A Problems/Problem 23

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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) \ }...

Solution

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.

See also

2006 AMC 12A (Problems • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
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