Deriving a formula for an integer sequence.

[Aion]

Can someone help me derive a formula for the following sequence, thanks!

[MATH]A, AB, -A(B^2),-A(B^3),A(B^4)...[/MATH]

 

[Aion]

I think I figured it out.

[MATH]a_{n}=A(B^{n-1})(-1)^{n}[/MATH]

Never mind, that does not work

 

[MarkFL]

It look to me like (if we are beginning at \(n=1\)):

[MATH]a_n=(-1)^{\left\lfloor\frac{n-1}{2}\right\rfloor}AB^{n-1}[/MATH]

 

[Aion]

I need to multiply [MATH]A(B^{n-1})[/MATH] with a function that outputs the series, [MATH]1,1,-1,-1,1 [/MATH].

 

[Aion]

It look to me like (if we are beginning at \(n=1\)):

[MATH]a_n=(-1)^{\left\lfloor\frac{n-1}{2}\right\rfloor}AB^{n-1}[/MATH]

I dont understand what [MATH]⌊(n−1)/2⌋ [/MATH] means :d. How did you find that fomula ?

 

[MarkFL]

That is the "floor" function, which rounds down to nearest integer. I used it to account for the fact that the sign appears to change on every other term rather than every term, which your formula does.

 

[Deleted member 4993]

How did you find that fomula ?

Mark probably had seen and solved 10000 of those problems...

 

[Aion]

Thanks I'll google the floor function =)

 

[pka]

I need to multiply [MATH]A(B^{n-1})[/MATH] with a function that outputs the series, [MATH]1,1,-1,-1,1 [/MATH].

That sequence could be the start in numerous sequences. SEE HERE

 

[Aion]

What about

[MATH]a_{n}=A(B^n)Sin(n\pi/2)+A(B^n)Cos(n\pi/2)[/MATH]

 

[pka]

What about [MATH]a_{n}=A(B^n)Sin(n\pi/2)+A(B^n)Cos(n\pi/2)[/MATH]

I thought it was more complicated. SEE HERE.
I was thinking of a programming assignment: \(1, 1,11,1,11,111,1,11,111,1111,\cdots\) to produce that string.