2008 Mock ARML 2 Problems/Problem 8

Problem

Given that $\sum_{i = 0}^{n}a_ia_{n - i} = 1$ and $a_n > 0$ for all non-negative integers $n$ , evaluate $\sum_{j = 0}^{\infty}\frac {a_j}{2^j}$ .

Solution

The motivating factor for this solution is the form of the first summation, which might remind us of the expansion of the coefficients of the product of two polynomials (or generating functions).

Let $x$ be an arbitrary number; note that

$\left[\sum_{j = 0}^{\infty} a_jx^j\right]^2 = (a_0 + a_1 \cdot x + a_2 \cdot x^2 + \cdots)^2\\ = a_0^2 + (a_0a_1 + a_1a_0)x + (a_0a_2 + a_1a_1 + a_2a_0)x^2 + \cdots$

By the given, the coefficients on the right-hand side are all equal to $1$ , yielding the geometric series:

$\left[\sum_{j = 0}^{\infty} a_jx^j\right]^2 = 1 + x + x^2 + \cdots = \frac{1}{1-x}$

For $x = \frac{1}{2}$ , this becomes $\left[\sum_{j = 0}^{\infty}\frac {a_j}{2^j}\right]^2 = 2$ , and the answer is $\boxed{\sqrt{2}}$ .

2008 Mock ARML 2 (Problems, Source)
Preceded by Problem 7	Followed by Final Question
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8

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2008 Mock ARML 2 Problems/Problem 8

Problem

Solution

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