- #1

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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]

- Thread starter gtfitzpatrick
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- #1

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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

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

- #3

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Thanks for any replies

- #4

Dick

Science Advisor

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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?

Thanks for any replies

- #5

Mark44

Mentor

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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

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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

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