Question

a12 and a20 are the 12th and 20th terms of Fibonacci sequence.find (a12,a20)

Answer 1

first i wrote the first terms of Fibonacci sequence.
1,1,2,3,5,8,13,21,34,55,89,144,233
then i solved these examples :
(a2,a4)=(1,3)=1=a2
(a3,a6)=(2,8)=2=a3
(a4,a6)=(3,8)=1=a2
so i guessed (am,an)=a(m,n)

Answer 2

what does: find (a12,a20) mean?

Answer 3

(a,b) means Greatest common divisor of a and b

Answer 4

hmm, you only wroted 13 terms .... any reason you didnt just compute it with the actual values?

Answer 5

1 = a1 or a2

Answer 6

not sure if there is a way i can think of to prove your conjecture tho ... unless we develop the closed form of a fibber

Answer 7

x^(n+2) = x^(n+1) + x^n

x^(n+2) - x^(n+1) - x^n = 0

x^n (x^2 - x - 1) = 0

x = 1/2 (1 +- sqrt(5))
.................................

setting a system we get

Ax1^0 + Bx2^0 = 1 , since f(0) = 1
Ax1^1 + Bx2^1 = 1 , since f(1) = 1
------------------

A + B = 1
Ax1 + Bx2 = 1


-Ax1 - Bx1 = -x1
Ax1 + Bx2 = 1
----------------
B(x2-x1) = 1-x1

B = (1-x1)/(x2-x1)

B = (1-(1 + sqrt(5))/2)/((1 - sqrt(5))/2-(1 + sqrt(5))/2)

B = (5-sqrt(5))/10

and A = 1- B

A = (5+sqrt(5))/10

the closed form is therefore:

Ax1^n + Bx2^n

Answer 8

of course im starting at n=0 .. so we would have to adjust for index

Answer 9

not sure if this would be useful in a GCD setup tho ... since its 2terms

Answer 10

well, it would at least be useful in defining the nth terms for large values of n, but then so could coding a computer script to work the manual stuff for you.