Question

The figures $F_1,$ $F_2,$ $F_3,$ and $F_4$ shown are the first in a sequence of figures. For $n \ge 3,$ $F_n$ 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 $F_{20}?$

$401$

$485$

$585$

$626$

$761$

Answer: E

Concepts:arithmetic sequence summation

Difficulty rating: 1630

Solution:

The outside square of $F_n$ has $4$ more diamonds than that of $F_{n-1},$ and the outside square of $F_2$ has $4,$ so the outside square of $F_n$ has $4(n - 1)$ diamonds.

Adding all the rings, $\begin{gathered} 1 + 4\big(1 + 2 + \cdots + (n - 1)\big) \\ = 1 + 4\cdot\frac{(n - 1)n}{2} \\ = 1 + 2(n - 1)n. \end{gathered}$

For $n = 20,$ this is $1 + 2\cdot 19\cdot 20 = 761.$

Thus, the correct answer is E.

2009 AMC 12A Problem 11

Answer: E

Difficulty rating: 1630

Solution:

Problem 11 in Other Years