2009 AMC 10A Problem 15

1. Count the first figures: $F_1 = 1$, $F_2 = 5$, $F_3 = 13$, $F_4 = 25$. The jumps are $4, 8, 12$.
2. See where the jumps come from. The new outside square of $F_n$ has $n$ diamonds on each side. Four sides of $n$ would be $4n$, but the $4$ corner diamonds are each counted twice, so the ring holds $4n - 4 = 4(n - 1)$ diamonds. For $n = 3$ that is $8$, for $n = 4$ it is $12$. This matches.
3. So each figure is the one before plus $4(n-1)$, and
$$F_{20} = 1 + 4(1) + 4(2) + 4(3) + \cdots + 4(19) = 1 + 4(1 + 2 + \cdots + 19).$$
4. The sum $1 + 2 + \cdots + 19$ is $\dfrac{19 \times 20}{2} = 190$.
5. So $F_{20} = 1 + 4 \times 190 = 761$.
6. Check the method on a figure we know: $F_4 = 1 + 4(1 + 2 + 3) = 1 + 24 = 25$. Correct.

The answer is $\boxed{\textbf{(E)}\ 761}$.

When a figure grows by adding a layer each time, count what one layer adds, then add up the layers. Here the layers add $4, 8, 12, \ldots$, so the total uses the sum $1 + 2 + \cdots + n$. Always test your method on the figures the problem shows you before using it for $n = 20$.

Look at the diamonds in rows along the diagonals. $F_n$ is an $n$ by $n$ grid of diamonds with an $(n-1)$ by $(n-1)$ grid fitted into the gaps: $F_3 = 9 + 4 = 13$ and $F_4 = 16 + 9 = 25$. So $F_{20} = 20^2 + 19^2 = 400 + 361 = 761$.

Count the diamonds in the first figures: $F_1 = 1$, $F_2 = 5$, $F_3 = 13$, $F_4 = 25$. The increases are $4, 8, 12$.

Why: the outer square of $F_n$ has $n$ diamonds on each side, and a square with $n$ diamonds per side contains $4n - 4 = 4(n-1)$ diamonds (four corners are shared). So
$$F_n = F_{n-1} + 4(n-1).$$

Summing the increments from $F_1 = 1$:
$$F_n = 1 + 4\big(1 + 2 + \cdots + (n-1)\big) = 1 + 4 \cdot \frac{(n-1)n}{2} = 2n(n-1) + 1.$$

Check: $F_3 = 2 \cdot 3 \cdot 2 + 1 = 13$ and $F_4 = 2 \cdot 4 \cdot 3 + 1 = 25$, matching the figure.

Then $F_{20} = 2 \cdot 20 \cdot 19 + 1 = 760 + 1 = 761$.

The answer is $\boxed{\textbf{(E)}\ 761}$.