1) This document discusses calculating Fibonacci numbers using a closed-form formula and taking the limit as n approaches infinity. It derives the formulas Fn = φn/√5 and ln(φ1000/5) = 1000lnφ - ln5/2. 2) It then calculates the value of φ to high precision and uses the formulas to find the limit of Fn as n approaches infinity, which is approximately 209. 3) Additional comments provide context that this relates to efficiently calculating Fibonacci numbers using memoization and discuss applications in computer science.

1. 1.2.8
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2. 1[10].
? ,1 ?
1 = 2 = F3
F14 = 377

3. 2[20]. (15) , F1000
?( A
). -_-
√
(15) Fn = φn / 5
? ?
√ 1
l n (φ 1000
/ 5) = 1000 l n φ − l n 5
2

4. 10
lnφ = 0.4812118....
ln5 = 1.60944....
1000 lnφ = 481.2118...
1
ln 5 = 0.80472
2
1
1000 lnφ − ln 5 = 480.40711
2

5. 2 . ?
log10 F1000 ?
1/(ln10)
208.64
F1000
4 209 .

6. 3[25]. F1 F1000 10
( 2 .)

7. ?
http://ohyecloudy.com/pnotes/archives/497
(memoization)
.
.
.

9. ..
.. 209 ...
http://bytes.com/topic/c/answers/63014-fibonacci-numbers
...

12. 4[14] n .
Fn = n