Sequences Series geometric series or an arithmetic series?

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physics246
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This is the sequence: 1, 2, 5, 14, 41, 122

1. Is this a geometric series or an arithmetic series?
2. I know the formula is a sub n=[3^(n-1)+1]/2, but how do you get that from a sub n=a sub 1 * r^(n-1), which is the geometric formula for series.
 
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So how would you get it then, with what formula or method?
 
Well, if you are clever, you will see that if n>=2, then we have:

[tex]a_{n}-3a_{n-1}=-1[/tex]

Assuming this is the pattern for all the next numbers, you may derive that explicit formula.
 
So there is no definite way of determining that formula? I see that to get from a term to another you add first 1, then 3, then 9, then 27, then 81 which are multiples of three. Does this have anything to do with determining the formula?

Thanks
 
A pattern that fits this is: [tex]F(N)=\frac{3^N+1}{2}, N=0,1,2..[/tex]
 
I know that that is the pattern, but I was just wondering how to figure that out with a formula or something.
 
1)neither

2) you can convert it to geometric series >>

t=1+2+5+14+41+122... Tn ------(1)
t= 1+2+5+14+41+122...Tn-Tn-1+Tn ------(2)

now eq 1- 2 and you'll get geometric series .
 
Can you please explain that more?
 
A finite sequence can be extended in infinitely many ways, i.e, there exists an infinity of patterns to choose from.

And yet, in the context of a math education, only one of these is considered to have a "formula". Although we with our modern sensibilities abhor this notion of formula, Euler would concur.
 
physics246 said:
Can you please explain that more?

ok

Tn=1+2+5+14+41+122... Tn ------(1)
Tn=0+1+2+5+14+41+122...Tn-1+Tn ------(2)

-------------------------------------------------------------- Eq 1- 2

0 = 1+1+3+3^2+3^3+3^4+... (Tn-Tn-1) - Tn

now transfer that Tn to that side (where zero is) & other side will have (n-1) terms and if you'll not include 1 of (first one) of series then terms will be (n-2) .that fact is that when you get such type of serieses you have to see for diffrences of series .

:devil:
 
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Crosson said:
And yet, in the context of a math education, only one of these is considered to have a "formula".


Usualy, assuming that we deal with a homogeneous linear sequence, the recurrence relation which we have to seek is that having the smallest degree, in this case a[k]-4a[k-1]+3a[k-2]=0 which gives immediately the closed form from the OP (k=3,4...).

[The characteristic equation is r^2-4*r+3=0 ---> r1=1; r2=3

Therefore we must seek a solution of the form a[k]=A*(1)^k+B*[(3)^k] (1); A,B = constants

we have a[k=3]=5 and a[k=4]=14 ---> replacing k in (1) with 3 and 4 results a system of equations from which A=1/2 and B=1/6.]
 
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