what is d 1200th number in the fibonacci sequence?

Answer 1

Try this:
http://mathforum.org/library/drmath/view/52690.html

Answer 2

Let Fn be the nth term in the Fibonacci sequence
and Ln the nth term in the Lucas sequence.
The main idea here is to give a method
to compute F1200 from smaller terms in the
sequences Fn and Ln.
If we know F12, F13, L12 and L13 we can find F1200.
We use the following identities:
F(n+m) = ½(FnLm + LnFm) (1)
L(n+m) = ½(LnLm + 5FmFn). (2)
We also need the duplication formulas:
L2n = Ln²-2(-1)^n
F2n = Ln*Fn
Plan of attack: Note that 1200 = 75*16.
So it is enough to find L75 and F75 and then
use the duplication formula 4 times to
get F1200.
So we have F12 = 144, F13 = 233,
L12 = 322, L13 = 521
and we get
L25 167761
F25 75025
from (1) and (2).
Now the duplication formulas give
L50 28143753123
F50 12586269025
Applying (1) and (2) once more yields
L75 4721424167835364
F75 2111485037978050
Finally, applying the duplication formula 4 more times,
we get the desired result:
F1200 =
27269884455406270157991615
31364219870500077999291772
58211805028949747264763730
26809482509284562310031170
17238012762721449359761674
38564430160399722058474059
17634660750474914561879656
76326865852809219571562607
32482240677942538091322190
56382939163918400

Answer 3

Here's the formula.

1/√5[((1 + √5)/2)^1200) - ((1 - √5)/2)^1200)]

Good luck in computing this number. It is absolutely enormous!

By the way. You don't mean "What is the last digit of the 1200th number" do you?

If yes, search the archives. That question was answered a few days ago.

Answer 4

give me a couple of mins...


...OK, its
2.72698844554062701579916153
1364219870500077992917725821
1805028949747264763730268094
8250928456231003117017238012
7627214493597616743856443016
x 10ú251

Unless I have made a typo, but I looked at it about ten times to make sure. =D

Answer 5

wow! unless there is a formula to figure this out this would take forever. sorry but I don't know what to tell you.