# Can't figure out this sequence.

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.

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