Question

For $-1 \lt r \lt 1,$ let $S(r)$ denote the sum of the geometric series $12 + 12r + 12r^2 + 12r^3 + \cdots.$ Let $a$ between $-1$ and $1$ satisfy $S(a)S(-a) = 2016.$ Find $S(a) + S(-a).$

Answer

Answer: 336

Concepts:geometric sequence algebraic manipulation

Difficulty rating: 1840

Solution:

The geometric series sums to $S(r) = \frac{12}{1 - r}.$ Therefore $\begin{aligned} S(a)S(-a) &= \frac{12}{1 - a} \cdot \frac{12}{1 + a} \\ &= \frac{144}{1 - a^2} = 2016, \end{aligned}$ so $\frac{1}{1 - a^2} = 14.$

Adding the two sums over a common denominator, $\begin{aligned} S(a) + S(-a) &= \frac{12}{1 - a} \\ &\quad {}+ \frac{12}{1 + a} \\ &= \frac{24}{1 - a^2} \\ &= 24 \cdot 14 = 336. \end{aligned}$

2016 AIME I Problem 1

Answer: 336

Difficulty rating: 1840

Solution:

Problem 1 in Other Years