2024 AMC 10B Problem 23

P21-25~6 minPrint

Sequences & Series Algebraic Manipulation Set up the equation/formula and compute; no special trick needed Compute small cases, spot the pattern, generalize

The Fibonacci numbers are defined by $F_1 = 1, F_2 = 1,$ and $F_n = F_{n-1} + F_{n-2}$ for $n \geq 3.$ What is ${\frac{F_2}{F_1}} + {\frac{F_4}{F_2}} + {\frac{F_6}{F_3}} + ... + {\frac{F_{20}}{F_{10}}}?$

Show solution without answering

Key insight

The ten fractions come out to whole numbers 1, 3, 4, 7, 11, ... and each one is the sum of the two before it, just like Fibonacci numbers.

Solution

List the Fibonacci numbers up to $F_{20}$ . Each is the sum of the two before it:

$n$	1	2	3	4	5	6	7	8	9	10
$F_n$	1	1	2	3	5	8	13	21	34	55

$n$	11	12	13	14	15	16	17	18	19	20
$F_n$	89	144	233	377	610	987	1597	2584	4181	6765

2. Work out the first few fractions.
$\frac{F_2}{F_1} = \frac11 = 1, \quad \frac{F_4}{F_2} = \frac31 = 3, \quad \frac{F_6}{F_3} = \frac82 = 4, \quad \frac{F_8}{F_4} = \frac{21}{3} = 7, \quad \frac{F_{10}}{F_5} = \frac{55}{5} = 11.$
3. Spot the pattern. $1 + 3 = 4$ , $3 + 4 = 7$ , $4 + 7 = 11$ . Each answer is the sum of the two before it, just like the Fibonacci rule, but starting $1, 3$ .
4. Continue the pattern, and check it against the table.
- Next: $7 + 11 = 18$ . Check: $\frac{F_{12}}{F_6} = \frac{144}{8} = 18$ .
- $11 + 18 = 29$ . Check: $\frac{377}{13} = 29$ .
- $18 + 29 = 47$ . Check: $\frac{987}{21} = 47$ .
- $29 + 47 = 76$ . Check: $\frac{2584}{34} = 76$ .
- $47 + 76 = 123$ . Check: $\frac{6765}{55} = 123$ .
5. Add the ten numbers.
$1 + 3 + 4 + 7 + 11 + 18 + 29 + 47 + 76 + 123 = 319.$

The answer is $\boxed{\textbf{(B)}\ 319}$ .

Why this works

The fractions $\frac{F_{2n}}{F_n}$ form a second Fibonacci-style sequence that starts $1, 3$ (these are called the Lucas numbers). When a problem builds new numbers out of Fibonacci numbers, work out a few and test whether they follow the same "add the last two" rule. If they do, the rest is small additions, and dividing the big numbers becomes a check rather than the main work.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

List the first Fibonacci numbers: $1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, \dots$

Compute the first few ratios $\frac{F_{2n}}{F_n}$ :
$\frac{F_2}{F_1} = 1,\quad \frac{F_4}{F_2} = 3,\quad \frac{F_6}{F_3} = \frac{8}{2} = 4,\quad \frac{F_8}{F_4} = \frac{21}{3} = 7,\quad \frac{F_{10}}{F_5} = \frac{55}{5} = 11.$
These are $1, 3, 4, 7, 11$ : each term is the sum of the previous two. This is no accident. The identity $F_{2n} = F_n(F_{n-1} + F_{n+1})$ shows $\frac{F_{2n}}{F_n} = F_{n-1} + F_{n+1}$ , and a sum of two Fibonacci-type sequences satisfies the same recurrence. (Check: $F_{12}/F_6 = 144/8 = 18 = 7 + 11$ .)

So continue the recurrence to ten terms:
$1,\ 3,\ 4,\ 7,\ 11,\ 18,\ 29,\ 47,\ 76,\ 123.$
Their sum is $1 + 3 + 4 + 7 + 11 + 18 + 29 + 47 + 76 + 123 = 319$ .

The answer is $\boxed{\textbf{(B)}\ 319}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 21: Sequences & Series (§21.1, §21.2, §21.3, §21.4)
Introduction to Algebra, Chapter 11: Special Factorizations (§11.2, §11.3, §11.5)
Prealgebra, Chapter 15: Problem-Solving Strategies (§15.1)
Intermediate Algebra, Chapter 17: More Sequences and Series (§17.1, §17.2)