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So far I have this relation of recurrence: $$a_n=-a_{n-2}$$ With $a_1=1$ and $a_2=2$.

I know I shoud replace $C \times r^n $ for $a_n$ and $C \times r^{n-2}$ for $a_{n-2}$

Then the relation gets here: $$Cr^{n}+Cr^{n-2}=0$$

If I divide the left side by $Cr^{n-2}$ The equation will be like this $r^2+1=0$, thus $r=-i$ or $r=i$

After this a can't solve it because I forgot complex numbers Can you guys please help me?

Axel
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Johnnuy junk
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  • If you write down the first few terms, the pattern should be very obvious. Hint: $a_n \ne cr^n\ \forall c,r$. – player3236 Dec 19 '20 at 08:39

2 Answers2

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Your sequence is $$1,2,-1,-2,1,2,-1,-2,....$$

So $$a_n =\cases{(-1)^{n-1\over 2}& if n odd \\ -2(-1)^{n\over 2}&if n even }$$

Or, as you put, $$a_n = ai^n +b(-i)^n$$ for some complex $a$ and $b$. From boundary values we have \begin{align}ai-bi &= 1\\ -a-b&=2\end{align}

So $a= {1-2i\over 2}$ and $b={1+2i\over 2}$.

So $$\boxed{a_n = {1\over 2}\Big(i^n+(-i)^n\Big) -i\Big(i^n-(-i)^n\Big)}$$

nonuser
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So you have $a_r=Ai^r+B(-i)^r$ as a general solution and you want $a_1=1, a_2=2$ so $$a_1=Ai-Bi=1$$ and if we multiply this by $i$ we get $B-A=i$

And also $$a_2=Ai^2+B(-i)^2=2=-A-B$$so that $A+B=-2$

We add the two equations to get $2B=-2+i$ and subtract first from second to get $2A=-2-i$

I've just sketched this out to show that there is nothing to be afraid of - it is just simple arithmetic.

Mark Bennet
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