2000 AMC 10 Problem 12

The counts $1, 5, 13, 25$ have first differences $4, 8, 12$, which themselves go up by a constant $4$. Constant second differences mean the count is a quadratic in the figure number $n$.

Rather than solve for coefficients, match the data to a recognizable form:
$$1 = 0^2 + 1^2,\quad 5 = 1^2 + 2^2,\quad 13 = 2^2 + 3^2,\quad 25 = 3^2 + 4^2.$$
So figure $n$ contains $n^2 + (n+1)^2$ squares. As a check, the next difference should be $16$, and $4^2 + 5^2 = 41 = 25 + 16$.

For $n = 100$:
$$100^2 + 101^2 = 10000 + 10201 = 20201.$$

The answer is $\boxed{\textbf{(C)}\ 20201}$.

When first differences form an arithmetic sequence, the terms are quadratic, and small cases usually reveal the closed form faster than solving three equations. Here the closed form has a picture behind it: the diamond of figure $n$ is an $n \times n$ grid of squares interleaved with an $(n+1) \times (n+1)$ grid, rotated $45^\circ$. Always verify the guessed formula on one extra term before plugging in $100$.

Count directly: a central square plus four triangular arms, each holding $1 + 2 + \cdots + n$ squares. Figure $100$ has $1 + 4 \cdot \frac{100 \cdot 101}{2} = 1 + 20200 = 20201$.

• Off-by-one in the index: figures start at 0, so figure 100 is the 101st term; using 99^2 + 100^2 = 19801 gives (B).
• Summing the differences $4 + 8 + \cdots + 400 = 20200$ and forgetting to add the initial $1$.
• Extrapolating linearly from the first difference and getting something near $401$.