AMC12 2008 B

AMC12 2008 B · Q12

AMC12 2008 B · Q12. It mainly tests Sequences & recursion (algebra), Averages (mean).

For each positive integer $n$, the mean of the first $n$ terms of a sequence is $n$. What is the $2008$th term of the sequence?

对于每个正整数 $n$，该数列前 $n$ 项的平均数为 $n$。该数列的第 $2008$ 项是多少？

(A) 2008 2008

(B) 4015 4015

(D) 4,030,056 4,030,056

(E) 4,032,064 4,032,064

Answer

Correct choice: (B)

正确答案：（B）

Solution

Letting $S_n$ be the nth partial sum of the sequence: $\frac{S_n}{n} = n$ $S_n = n^2$ The only possible sequence with this result is the sequence of odd integers. $a_n = 2n - 1$ $a_{2008} = 2(2008) - 1 = 4015 \Rightarrow \textbf{(B)}$

设 $S_n$ 为该数列的前 $n$ 项和： $\frac{S_n}{n} = n$ $S_n = n^2$ 满足这一结果的唯一可能数列是奇数数列。 $a_n = 2n - 1$ $a_{2008} = 2(2008) - 1 = 4015 \Rightarrow \textbf{(B)}$

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