Question

对每个正整数 $n$ ，某数列前 $n$ 项的平均数为 $n$ 。这个数列的第 $2008$ 项是多少？

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?

$2008$

$4015$

$4016$

$4{,}030{,}056$

$4{,}032{,}064$

因为前 $n$ 项平均数为 $n$ ，所以这些项之和为 $n^2$ 。

第 $n$ 项为 $n^2-(n-1)^2=2n-1$ ，所以第 $2008$ 项为 $2\cdot 2008-1=4015$ 。

所以正确答案是 B。

Since the mean of the first $n$ terms is $n,$ their sum is $n^2.$

The $n$ th term is $n^2-(n-1)^2=2n-1,$ so the $2008$ th term is $2\cdot 2008-1=4015.$

Thus, the correct answer is B.