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2009 AMC 10A Problem 15

2009 AMC 10A Problem 15AMC 10 Step by Step

Name: Date: September 13, 2026

Sequences & Series · difficulty 3 of 5 · about 4 min · position P11-15

The figures $F_1$ , $F_2$ , $F_3$ , and $F_4$ shown are the first in a sequence of figures. For $n\ge3$ , $F_n$ is constructed from $F_{n - 1}$ by surrounding it with a square and placing one more diamond on each side of the new square than $F_{n - 1}$ had on each side of its outside square. For example, figure $F_3$ has $13$ diamonds. How many diamonds are there in figure $F_{20}$ ?

A)
$401$
B)
$485$
C)
$585$
D)
$626$
E)
$761$

Answer

Key insight

Each new outer square adds 4(n-1) diamonds, so F_n = 1 + 4(1 + 2 + ... + (n-1)) = 2n(n-1) + 1, and F_20 = 761.

Solution

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 $\boxed{\textbf{(E)}\ 761}$ .

Why this works

A figure built by adding a layer each step is a sequence with a computable difference; when the differences grow linearly ( $4, 8, 12, \ldots$ ), the total is quadratic and a partial sum of an arithmetic series gives the closed form. Always test the formula on the values the problem hands you before evaluating at $n = 20$ .

Alternative approach

$F_n = n^2 + (n-1)^2$ : the diamonds form a diagonal grid consisting of an $n \times n$ array interleaved with an $(n-1) \times (n-1)$ array (the centered square numbers $1, 5, 13, 25, \ldots$ ). Then $F_{20} = 400 + 361 = 761$ .

The trap

Guessing a formula like n^2 (giving 400) from too few terms instead of verifying against 1, 5, 13, 25 and the stated F_3 = 13.

Common mistakes

Guessing a formula like n^2 (giving 400) from too few terms instead of verifying against 1, 5, 13, 25 and the stated F_3 = 13.
Counting $4n$ diamonds on the new square (double-counting the corners), which gives $F_n = 2n^2 + 2n - 3$ and a wrong total.

Techniques

Compute small cases, spot the pattern, generalize