-1
$\begingroup$

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.

$\endgroup$
3
  • 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
1
$\begingroup$

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 $c_1,c_2$.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.