AMC10 2008 B

AMC10 2008 B · Q13

AMC10 2008 B · Q13. It mainly tests Sequences & recursion (algebra).

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

对于每个正整数 $n$，序列前 $n$ 项的平均值为 $n$。求该序列的第 2008 项。

(A) 2008 2008

(B) 4015 4015

(D) 4,030,056 4030056

(E) 4,032,064 4032064

Answer

Correct choice: (B)

正确答案：（B）

Solution

Because the mean of the first $n$ terms is $n$, their sum is $n^2$. Therefore the $n$th term is $n^2 - (n-1)^2 = 2n - 1$, and the 2008th term is $2 \cdot 2008 - 1 = 4015$.

因为前 $n$ 项的平均值为 $n$，它们的和为 $n^2$。因此第 $n$ 项为 $n^2 - (n-1)^2 = 2n - 1$，第 2008 项为 $2 \cdot 2008 - 1 = 4015$。

Topics

AL.012 Sequences & recursion (algebra)