2008 AMC 10B Problems/Problem 13

Problem

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?

Since the mean of the first $n$ terms is $n$ , the sum of the first $n$ terms is $n^2$ . Thus, the sum of the first $2007$ terms is $2007^2$ and the sum of the first $2008$ terms is $2008^2$ . Hence, the 2008th term is $2008^2-2007^2=(2008+2007)(2008-2007)=4015\Rightarrow \boxed{B}$

Note that $n^2$ is the sum of the first n odd numbers.