2009 AMC 12A Problem 11

Problem 11 of 25IntermediateAlgebra

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}?

Show solution

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.