I can't figure this one out, I'm all out of brains.

aₙ = aₙ₋₁ + aₙ₋₂

a₁ = 1; a₂ = -2

I need to find the 4th and the 10th numbers in the sequence and apparently the answers are a₄ = -3 and a₁₀ = -47. I tried an equation system and everything I know on this problem but I can't get those answers. I've never seen a sequence like that and I don't see a common difference here.

I would be very thankful for any help.

  • 3
    $\begingroup$ I don't understand the issue here. You just use the supplied formula $8$ times. $\endgroup$ – Peter Foreman May 13 '19 at 17:26
  • $\begingroup$ Do you mean $a_n=a_{n-1}+a_{n-2}$? $\endgroup$ – Thomas Andrews May 13 '19 at 17:29
  • $\begingroup$ Make the ansatz $$a_n=q^n$$ $\endgroup$ – Dr. Sonnhard Graubner May 13 '19 at 17:30

You have $a_n-a_{n-1}-a_{n-2}=0$. Let the solution be $a_n=ck^n,k\ne0,c\in\Bbb R$. Substituting in the equation,$$k^2-k-1=0\\\therefore k=\frac{1\pm\sqrt5}2$$The general solution of the recurrence is$$a_n=c_1\left(\frac{1+\sqrt5}2\right)^n+c_2\left(\frac{1-\sqrt5}2\right)^n$$Now use the values provided to find $c_1,c_2$.


Let's first calculate $a_3=a_2+a_1$ now $a_4=a_3+a_2=2.a_2+a_1$, $a_5=a_4+a_3=3a2+2a_1$ therefore we can conclude that $a_n=F_{n-1}.a_2+F_{n-2}.a_1$ where F_n is the $n^{th}$fibonacci number. Note:- when $a_2,a_1=1$ then the sequence is fibonacci itself.


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