2008 AMC 10B Problem 13

An average tells you a total: total $=$ average $\times$ how many. So turn the averages into running totals, then find each term by subtracting.

1. The first $1$ term has average $1$, so it adds to $1 \times 1 = 1$.
2. The first $2$ terms have average $2$, so they add to $2 \times 2 = 4$.
3. The first $3$ terms have average $3$, so they add to $3 \times 3 = 9$.
4. Each term is how much the running total grows when that term is added:

1. The terms are the odd numbers $1, 3, 5, 7, \ldots$ The term in position $n$ is $2n - 1$.
2. So the $2008$th term is $2 \times 2008 - 1 = 4015$.

The answer is $\boxed{\textbf{(B)}\ 4015}$.

A question about averages of "the first $n$ terms" is really about running totals, and any single term is the difference between two running totals next to each other. Here the running totals are the perfect squares, and the gaps between neighboring squares are the odd numbers. That is the same fact as "the first $n$ odd numbers add up to $n \times n$".

Subtract the two running totals directly. The first $2008$ terms add to $2008^2$ and the first $2007$ add to $2007^2$, so the $2008$th term is
$$2008^2 - 2007^2 = (2008 - 2007)(2008 + 2007) = 1 \times 4015 = 4015,$$
using the difference of squares $A^2 - B^2 = (A - B)(A + B)$.

Let $S_n$ be the sum of the first $n$ terms. A mean of $n$ over $n$ terms means
$$S_n = n \cdot n = n^2.$$
The $2008$th term is the amount the sum grows when that term is added:
$$a_{2008} = S_{2008} - S_{2007} = 2008^2 - 2007^2 = (2008 - 2007)(2008 + 2007) = 4015.$$

The answer is $\boxed{\textbf{(B)}\ 4015}$.