HMMT November

Algebra Difficulty 5.1 AIME, harder Prove it United States

Problem:

Let $x < 0.1$ be a positive real number. Let the foury series be $4 + 4x + 4x^{2} + 4x^{3} + \ldots$ , and let the fourier series be $4 + 44x + 444x^{2} + 4444x^{3} + \ldots$ . Suppose that the sum of the fourier series is four times the sum of the foury series. Compute $x$ .

Solutions — 2

Solution 1

Solution:

The sum of the foury series can be expressed as $\frac{4}{1-x}$ by geometric series. The fourier series can be expressed as
$\begin{aligned} & \frac{4}{9}\left((10-1)+(100-1)x+(1000-1)x^{2}+\ldots\right) \\ & = \frac{4}{9}\left(\left(10+100x+1000x^{2}+\ldots\right)-\left(1+x+x^{2}+\ldots\right)\right) \\ & = \frac{4}{9}\left(\frac{10}{1-10x}-\frac{1}{1-x}\right) \end{aligned}$
Now we solve for $x$ in the equation
$\frac{4}{9}\left(\frac{10}{1-10x}-\frac{1}{1-x}\right) = 4 \cdot \frac{4}{1-x}$
by multiplying both sides by $(1-10x)(1-x)$ . We get $x = \frac{3}{40}$ .

Solution 2

Solution:

Let $R$ be the sum of the fourier series. Then the sum of the foury series is $(1-10x)R$ . Thus, $1-10x = 1/4 \Longrightarrow x = 3/40$ .

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