Mock AIME 4 Pre 2005/Problems/Problem 9

Problem

The value of the sum $\sum_{n=1}^\infty\frac{(7n+32)\cdot3^n}{n\cdot(n+2)\cdot4^n}$ can be expressed in the form $\tfrac pq$ , for some relatively prime positive integers $p$ and $q$ . Compute the value of $p+q$ .

Solution

Intuition: the infinite sum converges as it decreases "sub-geometrically", and we hope that the exact value can be computed with some sort of telescoping terms.

Letting the desired sum be

$S = \sum^{\infty}_{n=1} \frac{(7n + 32) \cdot 3^n}{n \cdot (n + 2) \cdot 4^n}$

we know we can work with the "geometric term" easily as we can shift the terms by multiplying $S$ by any power of $\frac{3}{4}$ .

Now, we take a look at the remaining part and see if it can be rewritten as partial fractions:

$\frac{7n + 32}{n \cdot (n + 2)} = \frac{A}{n} - \frac{B}{n + 2}$

and indeed we must have $A - B = 7$ , and $2A = 32$ , yielding $A = 16$ and $B = 9$ . The fact that the ratio between $B$ and $A$ is precisely $\left (\frac{3}{4} \right )^2$ is very encouraging.

Now we rewrite the sum as

$\begin{align*} S &= \sum^{\infty}_{n=1} \frac{3^n}{4^n} \left ( \frac{16}{n} - \frac{9}{n + 2} \right ) \\ &= \sum^{\infty}_{n=1} \frac{3^n}{4^n} \frac{16}{n} - \sum^{\infty}_{n=1} \frac{3^n}{4^n} \frac{9}{n + 2} \\ &= \left ( \frac{3}{4} \frac{16}{1} + \frac{9}{16} \frac{16}{2} + \sum^{\infty}_{n=3} \frac{3^n}{4^n} \frac{16}{n} \right )- \sum^{\infty}_{n=1} \frac{3^n} {4^n} \frac{9}{n + 2} \\ &= \left ( \frac{33}{2} + \sum^{\infty}_{n=1} \frac{3^2\cdot3^n}{4^2\cdot 4^n} \frac{16}{n + 2} \right )- \sum^{\infty}_{n=1} \frac{3^n} {4^n} \frac{9}{n + 2} \\ &= \frac{33}{2} + \sum^{\infty}_{n=1} \frac{3^n}{4^n} \frac{9}{n + 2} - \sum^{\infty}_{n=1} \frac{3^n} {4^n} \frac{9}{n + 2} \\ &= \frac{33}{2} \end{align*}$

With that, our desired answer is $33 + 2 = \fbox{35}$

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Mock AIME 4 Pre 2005/Problems/Problem 9

Problem

Solution