AMC10 2022 B

AMC10 2022 B · Q15

AMC10 2022 B · Q15. It mainly tests Sequences & recursion (algebra).

Let $S_n$ be the sum of the first $n$ terms of an arithmetic sequence that has a common difference of $2$. The quotient $\frac{S_{3n}}{S_n}$ does not depend on $n$. What is $S_{20}$?

设$S_n$是一个公差为$2$的等差数列的前$n$项和。商$\frac{S_{3n}}{S_n}$不依赖于$n$。求$S_{20}$。

(A) 340 340

(B) 360 360

(D) 400 400

(E) 420 420

Answer

Correct choice: (D)

正确答案：（D）

Solution

Let's say that our sequence is \[a, a+2, a+4, a+6, a+8, a+10, \ldots.\] Then, since the value of n doesn't matter in the quotient $\frac{S_{3n}}{S_n}$, we can say that \[\frac{S_{3}}{S_1} = \frac{S_{6}}{S_2}.\] Simplifying, we get $\frac{3a+6}{a}=\frac{6a+30}{2a+2}$, from which \[\frac{3a+6}{a}=\frac{3a+15}{a+1}.\] \[3a^2+9a+6=3a^2+15a\] \[6a=6\] Solving for $a$, we get that $a=1$. Since the sum of the first $n$ odd numbers is $n^2$, $S_{20} = 20^2 = \boxed{\textbf{(D) } 400}$. Note: you could also plug in the formulas for $\frac{S_{3n}}{S_{n}}$ and simplify, getting $3+ \frac{6n}{a+n-1}$ You would then find a=$1$

设数列为\[a, a+2, a+4, a+6, a+8, a+10, \ldots.\] 由于商$\frac{S_{3n}}{S_n}$不依赖$n$，可设\[\frac{S_{3}}{S_1} = \frac{S_{6}}{S_2}.\] 化简得$\frac{3a+6}{a}=\frac{6a+30}{2a+2}$，即\[\frac{3a+6}{a}=\frac{3a+15}{a+1}.\] \[3a^2+9a+6=3a^2+15a\] \[6a=6\] 解得$a=1$。前$n$个奇数和为$n^2$，故$S_{20} = 20^2 = \boxed{\textbf{(D) } 400}$。注：也可代入$S_{3n}/S_n$公式化简，得$3+ \frac{6n}{a+n-1}$，从而求$a=1$。

Topics

AL.012 Sequences & recursion (algebra)