AIME 2016 I · 第 1 题

AIME 2016 I — Problem 1

专题: Contest Math
难度: L4
来源: AIME

题目详情

Problem 1

For $-1, let$ 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 $S(a)+S(-a)$ .

解析

Solution 1

The sum of an infinite geometric series is $\frac{a}{1-r}\rightarrow \frac{12}{1\mp a}$ . The product $S(a)S(-a)=\frac{144}{1-a^2}=2016$ . $\frac{12}{1-a}+\frac{12}{1+a}=\frac{24}{1-a^2}$ , so the answer is $\frac{2016}{6}=\boxed{336}$ .

Solution 2

There is a clear infinite geometric series, so we use the geometric series formula $S = \frac{a}{1-r}$ . However, adjusted for the context of this specific series, $S(a)=\frac{12}{1-a}$ .

Plugging in the formula for both sum and product results in $\frac{12}{1-a} \times \frac{12}{1+a} = 2016$ for the product, and $\frac{12}{1-a} + \frac{12}{1+a}$ for the sum. This ensures both fractions have the same denominator, so that solving would be easier to work with.

Upon first notice, dividing by $(1-a)(1+a)$ is equivalent to multiplying by 14.

Unsimping the top also results in a constant, as $12(1+a)+12(1-a)=12 + 12a + 12 - 12a = 24 + 0a = 24$ .

We can use the fact that dividing by $(1-a)(1+a)$ is equivalent to multiplying by 14 in this scenario, as we have a constant term, 24, in the numerator of the second fraction. Multiplying this by 14 results in $24 \times 14 = \boxed{336}$ .

~AlgowheelAZ1

Video Solution by OmegaLearn

https://youtu.be/3wNLfRyRrMo?t=153

~ pi_is_3.14