2024 AMC 10B Problem 23

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 $\boxed{\textbf{(B)}\ 319}$.

The ratios $F_{2n}/F_n$ are the Lucas numbers $L_n = F_{n-1} + F_{n+1}$, a second Fibonacci-like sequence with starting values $1, 3$. Whenever a problem builds a new sequence out of Fibonacci numbers, compute a handful of terms and test whether they obey the same recurrence; if so, the whole problem reduces to small additions. The identity behind it, $F_{2n} = F_n L_n$, is worth remembering.

Brute force: extend the list to $F_{20} = 6765$ and divide each even-indexed term by the corresponding term ($F_{14}/F_7 = 377/13 = 29$, $F_{16}/F_8 = 987/21 = 47$, $F_{18}/F_9 = 2584/34 = 76$, $F_{20}/F_{10} = 6765/55 = 123$). Same ten numbers, sum $319$.

• Arithmetic slips while computing Fibonacci numbers up to F_20 = 6765 and dividing; the recurrence on the ratios avoids all large numbers.
• Starting the ratio sequence at $3, 4, 7, \dots$ and dropping the first term $F_2/F_1 = 1$, giving $318$.