- #1

- 380

- 0

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]

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter gtfitzpatrick
- Start date

- #1

- 380

- 0

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]

- #2

- 380

- 0

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

- #3

- 380

- 0

Thanks for any replies

- #4

Dick

Science Advisor

Homework Helper

- 26,263

- 619

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

Mark44

Mentor

- 36,031

- 7,961

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 a

- #6

- 116

- 0

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 a

Share: