2000 AMC 10 Problem 12

1. Look at the counts: $1, 5, 13, 25$. They grow by $4, 8, 12$, and those jumps grow by $4$ each time. So the next jump is $16$ and figure $4$ has $41$ squares.
2. Try to spot a pattern using square numbers:
$$1 = 0^2 + 1^2,\quad 5 = 1^2 + 2^2,\quad 13 = 2^2 + 3^2,\quad 25 = 3^2 + 4^2.$$
Figure $n$ seems to have $n^2 + (n+1)^2$ squares.
3. Check it on figure $4$, which we found has $41$ squares: $4^2 + 5^2 = 16 + 25 = 41$. It matches.
4. The pattern also matches the picture. Read figure $n$ row by row: the rows have $1, 3, 5, \ldots, 2n+1, \ldots, 5, 3, 1$ squares. The top part down to the longest row is the first $n+1$ odd numbers, which add to $(n+1)^2$. The part below is the first $n$ odd numbers, which add to $n^2$. For figure $2$: $1 + 3 + 5 = 9$ and $3 + 1 = 4$, total $13$.
5. For figure $100$, use $n = 100$ (not $99$, since the first figure is figure $0$):
$$100^2 + 101^2 = 10000 + 10201 = 20201.$$

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

When the jumps between terms themselves grow by the same amount each time, the pattern is a "squared" formula, and small cases usually show it quickly. Always check a guessed formula on one more term before using it for $n = 100$, and watch whether the list starts at $0$ or $1$.

Count the picture directly. Each figure has one center square and four arms. In figure $n$, each arm is a staircase with $1 + 2 + \cdots + n$ squares. For $n = 100$, one arm has $\frac{100 \cdot 101}{2} = 5050$ squares, so the total is $1 + 4 \cdot 5050 = 20201$.