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 $n$ 项的平均数为 $n$ ，所以它们的和为 $n \cdot n = n^2$ 。

第 $n$ 项是相邻两个部分和的差， $n^2 - (n-1)^2 = 2n - 1$ 。

当 $n = 2008$ 时，该项为 $2 \cdot 2008 - 1 = 4015$ 。

因此，正确答案是 B。

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

The $n$ th term is the difference of consecutive sums, $n^2 - (n-1)^2 = 2n - 1.$

For $n = 2008,$ the term is $2 \cdot 2008 - 1 = 4015.$

Thus, the correct answer is B.