AMC12 2007 B

AMC12 2007 B · Q15

AMC12 2007 B · Q15. It mainly tests Sequences & recursion (algebra), Algebra misc.

The geometric series $a+ar+ar^2\ldots$ has a sum of $7$, and the terms involving odd powers of $r$ have a sum of $3$. What is $a+r$?

几何级数 $a+ar+ar^2\ldots$ 的和为 $7$，且含有 $r$ 的奇数次幂的项的和为 $3$。求 $a+r$。

(A) \frac{4}{3} \frac{4}{3}

(B) \frac{12}{7} \frac{12}{7}

(D) \frac{7}{3} \frac{7}{3}

(E) \frac{5}{2} \frac{5}{2}

Answer

Correct choice: (E)

正确答案：（E）

Solution

The sum of an infinite geometric series is given by $\frac{a}{1-r}$ where $a$ is the first term and $r$ is the common ratio. In this series, $\frac{a}{1-r} = 7$ The series with odd powers of $r$ is given as \[ar + ar^3 + ar^5 ...\] Its sum can be given by $\frac{ar}{1-r^2} = 3$ Doing a little algebra $ar = 3(1-r)(1+r)$ $ar = 3\left(\frac{a}{7}\right)(1+r)$ $\frac{7}{3}r = 1 + r$ $r = \frac{3}{4}$ $a = 7(1-r) = \frac{7}{4}$ $a + r =\boxed{ \frac{5}{2}} \Rightarrow \mathrm{(E)}$

无穷等比级数的和为 $\frac{a}{1-r}$，其中 $a$ 为首项，$r$ 为公比。在该级数中，$\frac{a}{1-r} = 7$ 含有 $r$ 的奇数次幂的级数为 \[ar + ar^3 + ar^5 ...\] 其和为 $\frac{ar}{1-r^2} = 3$ 做一些代数变形： $ar = 3(1-r)(1+r)$ $ar = 3\left(\frac{a}{7}\right)(1+r)$ $\frac{7}{3}r = 1 + r$ $r = \frac{3}{4}$ $a = 7(1-r) = \frac{7}{4}$ $a + r =\boxed{ \frac{5}{2}} \Rightarrow \mathrm{(E)}$

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