• Support PF! Buy your school textbooks, materials and every day products Here!

Fibonacci sequence

  • #1
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/[tex]\sqrt{5}[/tex][ ((1+[tex]\sqrt{}5[/tex])/2)^2 - ((1-[tex]\sqrt{}5[/tex])/2)^2]
 

Answers and Replies

  • #2
the first 6 terms are 1,1,2,3,5,8 but from here where do i go proving this... totally lost
 
  • #3
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
 
  • #4
Dick
Science Advisor
Homework Helper
26,258
618
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?
 
  • #5
33,162
4,846
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/[tex]\sqrt{5}[/tex][ ((1+[tex]\sqrt{}5[/tex])/2)^2 - ((1-[tex]\sqrt{}5[/tex])/2)^2]

This can't be the right formula for an. It varies with n, while your formula above is a constant.
 
  • #6
The characteristic equation is [tex] r^{2}-r-1=0[/tex]

The roots are [tex] \frac {1+ \sqrt{5}}{2} , \frac {1- \sqrt{5}}{2} [/tex]

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

Then use the initial conditions, namely a0=1 and a1=1 to find [tex] \alpha [/tex] and [tex] \beta [/tex]
 

Related Threads for: Fibonacci sequence

  • Last Post
Replies
13
Views
2K
  • Last Post
Replies
8
Views
1K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
2
Views
1K
Replies
6
Views
1K
Replies
7
Views
3K
  • Last Post
Replies
6
Views
1K
Top