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:recursion arithmetic sequence summation

Difficulty rating: 1400

Solution:

Going from $F_{n-1}$ to $F_n,$ 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.

2009 AMC 10A Problem 15

Answer: E

Difficulty rating: 1400

Solution:

Problem 15 in Other Years