Fibonacci sequence

[gtfitzpatrick]

the Fibonacci sequence is defined by

a1 = a2 = 1, a(n+2)] = an + a(n+1).

write out the first 6 terms of the sequence and prove that an = 1/$$\sqrt{5}$$[ ((1+$$\sqrt{}5$$)/2)^2 - ((1-$$\sqrt{}5$$)/2)^2]

 

[gtfitzpatrick]

the first 6 terms are 1,1,2,3,5,8 but from here where do i go proving this... totally lost

 

[gtfitzpatrick]

wait do i have to use the first 6 terms or do you think they just want me to prove it?
Thanks for any replies

 

[Dick]

wait do i have to use the first 6 terms or do you think they just want me to prove it?
Thanks for any replies

You've got the first 6 terms just fine. To find the general solution look for power law solutions an=r^n of your recursion relation. Put an=r^n into your recursion relation and solve for r. You should get a quadratic and two solutions. What are they?

 

[Mark44]

the Fibonacci sequence is defined by

a1 = a2 = 1, a(n+2)] = an + a(n+1).

write out the first 6 terms of the sequence and prove that an = 1/$$\sqrt{5}$$[ ((1+$$\sqrt{}5$$)/2)^2 - ((1-$$\sqrt{}5$$)/2)^2]

This can't be the right formula for a_n. It varies with n, while your formula above is a constant.

 

[Random Variable]

The characteristic equation is $$r^{2}-r-1=0$$

The roots are $$\frac {1+ \sqrt{5}}{2} , \frac {1- \sqrt{5}}{2}$$

The general solution is then $$a_{n} = \alpha (\frac {1+ \sqrt{5}}{2})^{n} + \beta ({\frac{1- \sqrt{5}}{2})^n$$

Then use the initial conditions, namely a₀=1 and a₁=1 to find $$\alpha$$ and $$\beta$$