# 2006 iTest Problems/Problem 36

Let $\alpha$ denote $\cos^{-1}(\tfrac 23)$. The recursive sequence $a_0,a_1,a_2,\ldots$ satisfies $a_0 = 1$ and, for all positive integers $n$, $$a_n = \dfrac{\cos(n\alpha) - (a_1a_{n-1} + \cdots + a_{n-1}a_1)}{2a_0}.$$ Suppose that the series $$\sum_{k=0}^\infty\dfrac{a_k}{2^k}$$ can be expressed uniquely as $\tfrac{p\sqrt q}r$, where $p$ and $r$ are coprime positive integers and $q$ is not divisible by the square of any prime. Find the value of $p+q+r$.

## Solution

We write $\sum_{k=0}^n a_k a_{n-k} = \cos(n\alpha)$ by rearranging the defining equation and using $a_0 = 1$. Summing this with weights $\frac{1}{2^n}$ from zero to infinity, we get $\sum_{n=0}^{\infty}\frac{1}{2^n}\sum_{n}^{k=0} a_k a_{n-k} = \sum_{n=0}^{\infty} \frac{\cos(n\alpha)}{2^n}$. We can rewrite this as $(\sum_{n=0}^{\infty} \frac{a_n}{2^n})^2 = \sum_{n=0}^{\infty} \frac{\cos(n\alpha)}{2^n}$.

Next, we compute$\sum_{n=0}^{\infty} \frac{\cos(n\alpha)}{2^n} = \frac{1}{2}(\sum_{n=0}^{\infty} \frac{e^{in\alpha}}{2^n} + \sum_{n=0}^{\infty} \frac{e^{-in\alpha}}{2^n}) = \frac{1}{2}(\frac{1}{1-\frac{e^{i\alpha}}{2}} + \frac{1}{1-\frac{e^{-i\alpha}}{2}})$, which simplifies to $\frac{1 - \frac{e^{i\alpha} + e^{-i\alpha}}{4}}{\frac{5}{4} - \frac{e^{i\alpha} + e^{-i\alpha}}{2}}$. Since $\frac{e^{i\alpha} + e^{-i\alpha}}{2} = \cos\alpha = \frac{2}{3}$, the entire expression becomes $\frac{1 - \frac{1}{2}\cdot\frac{2}{3}}{\frac{5}{4} - \frac{2}{3}} = \frac{\frac{2}{3}}{\frac{7}{12}} = \frac{8}{7}$.

Taking square roots, we get $\sum_{n=0}^{\infty} \frac{a_n}{2^n} = \sqrt{\frac{8}{7}} = \frac{2\sqrt{14}}{7}$, so our answer is $2 + 14 + 7 = \boxed{23}$ and we are done.