2009 AMC 10A Problem 15

Problem 15 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

Going from

F_{n-1}

F_n,

the new outside square carries

4(n-1)

diamonds. Starting from the single diamond of

F_1,

\begin{aligned} F_n &= 1 \\ &\quad {}+ 4\big(1 + 2 + \cdots + (n-1)\big) \\ &= 1 + 4 \cdot \dfrac{(n-1)n}{2} \\ &= 1 + 2n(n-1). \end{aligned}

Therefore

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

Thus, the correct answer is E.