# 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 $2008^{\text{th}}$ term of the sequence? $\mathrm{(A)}\ {{{2008}}} \qquad \mathrm{(B)}\ {{{4015}}} \qquad \mathrm{(C)}\ {{{4016}}} \qquad \mathrm{(D)}\ {{{4,030,056}}} \qquad \mathrm{(E)}\ {{{4,032,064}}}$

## Solution

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 $2008^2-2007^2=(2008+2007)(2008-2007)=4015\Rightarrow \boxed{\text{(B)}}$

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

## Solution 2 (Using Answer Choices)

From inspection, we see that the sum of the sequence is basically $n^2$. We also notice that $n^2$ Is the sum of the first $n$ ODD integers. Because $4015$ is the only odd integer, $\boxed{B}$ is the answer.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 