2006-07-25 07:55:03 · 9 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
fib(2012) = 1360339099050831291379435681688955503065585659010946155141602617183897337443701411096399942297131982079502638445774331157946277005087171888228558355537366224371746200469637880236316588599086887348036901884241593336618524538690660618063682125339544013281223536004712161112123160064091085502986418726938170867592022796613517119004637421258467517923413084049920657445698262851995845658288840175943184664815407670853363182269
I used some random web program, so I'm not garanteeing this to be correct. 2012 is also the end of the Mayan calendar.
2006-07-25 08:08:47 · answer #1 · answered by random.oracle_23 2 · 0⤊ 2⤋
One approach is to derive an expression for the nth term. This involves z-transforms -these are like discrete Laplace transforms. You get to learn about these in 2nd or 3rd year differential equations or in 1st year graduate school if you are working on digital signal processing. I worked this out once for fun.
The other approach is to write a short python program. This is recommended as it is freely available, has a great math library, etc.
Assuming that
term 1 is 1
term 2 is 1
term 3 is 2
then
term 2012 is 13603390990508312
91379435681688955503065585
659010946155141602617183
8973374437014110963
999422971319820795026384
4577433115794627700
5087171888228558355537
366224371746200469637
8802363165885990868873
480369018842415933366
18524538690660618063682
12533954401328122353
6004712161112123160
064091085502986418726938
1708675920227966135171
190046374212584675179
23413084049920657445698
26285199584565828884
017594318466481540767085
3363182269
======
Oooh, I like the excel idea. But, you probably won't get all the digits. Python doesn't drop any.
2006-07-25 08:20:57 · answer #2 · answered by Anonymous · 0⤊ 0⤋
The Fibonacci sequence is :
1 , 1 , 2 , 3 , 5 , 8 , 13 , 21, 34 , 55 , 89 . .............
its rule is :
S1 =1 , S2 = 1 , S(n+2) = S(n+1) + S(n)
So you an go so till you get S(2012)
Note : use Excel program to get it :
Put 1 in the cell A1 , 1 in A2 then in A3 enter = A1+A2
Then copy A3 to the cells from A4 till A2012
2006-07-25 08:12:06 · answer #3 · answered by a_ebnlhaitham 6 · 0⤊ 0⤋
in case you study the wikipedia link in C. Condap's positioned up, you'll stumble on a speedy thanks to compute the nth Fibonacci huge variety, quite: Compute ?^n / ?5. round your answer to the nearest total huge variety. The huge variety you get will honestly be the nth Fibonacci huge variety for any n > 0.
2016-11-25 23:24:07 · answer #4 · answered by drinkwater 4 · 0⤊ 0⤋
You can use the Binet formulas to find the number. At one point I had a maple program to find them.
http://milan.milanovic.org/math/english/relations/relation1.html
2006-07-25 08:03:20 · answer #5 · answered by raz 5 · 0⤊ 0⤋
Can I have 10 Guesses???
2006-07-25 07:59:16 · answer #6 · answered by Anonymous · 0⤊ 0⤋
3, always 3
2006-07-25 07:58:22 · answer #7 · answered by Anonymous · 0⤊ 0⤋
(1/(sqrt(5)))(((1 + sqrt(5))/2)^n - ((1 - sqrt(5))/2)^n)
(1/sqrt(5))(((1 + sqrt(5))/2)^2012 - ((1 - sqrt(5))/2)^(2012))
ANS : 1.36033909905 * 10^420
2006-07-25 15:26:43 · answer #8 · answered by Sherman81 6 · 0⤊ 0⤋
136033909905083129137943
568168895550306558565901
094615514160261718389733
744370141109639994229713
198207950263844577433115
794627700508717188822855
835553736622437174620046
963788023631658859908688
734803690188424159333661
852453869066061806368212
533954401328122353600471
216111212316006409108550
298641872693817086759202
279661351711900463742125
846751792341308404992065
744569826285199584565828
884017594318466481540767
0853363182269
2006-07-25 08:34:54 · answer #9 · answered by deflagrated 4 · 0⤊ 0⤋