To solve the problem of finding the sum of all even-valued terms in the Fibonacci sequence starting with 1 and 2, up to 4 million, follow these steps:
Key Observations
- The Fibonacci sequence starting with 1, 2 has parity (odd/even) pattern: odd, even, odd, odd, even, odd, odd, even...
- Even terms occur every third position (since odd+odd=even, breaking the cycle).
- Even terms follow a recurrence relation: (Ek = 4E{k-1} + E_{k-2}) (derived from the Fibonacci sequence's structure).
Recurrence for Even Terms
Let the even terms be (E_1=2, E_2=8, E_3=34, ...). Using the recurrence (Ek = 4E{k-1}+E_{k-2}), we compute even terms until they exceed 4 million, then sum them.
Calculation
- Start with (E_{prevprev}=2, E{prev}=8, sum=10).
- Compute next even terms:
(E{next}=4*8+2=34) → sum=44
(E{next}=434+8=144) → sum=188
(E_{next}=4144+34=610) → sum=798
(E{next}=4*610+144=2584) → sum=3382
(E{next}=42584+610=10946) → sum=14328
(E_{next}=410946+2584=46368) → sum=60696
(E{next}=4*46368+10946=196418) → sum=257114
(E{next}=4196418+46368=832040) → sum=1089154
(E_{next}=4832040+196418=3524578) → sum=4613732
Next term ((14930352)) exceeds 4 million → stop.
Answer: (\boxed{4613732})
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