What is the general formula for the terms in this sequence?

  • Thread starter Thread starter aruwin
  • Start date Start date
  • Tags Tags
    Notation
aruwin
Messages
204
Reaction score
0
I have a problem on how to generalize this sequence.
Problem:
In sequence a,b,x1,x2,...xn,...,each term is equals to the arithmetic mean of its two preceding numbers.Using a and b, find:

1.xn

2.lim(n->∞) xn

My working:

x1 = (a+b)/2, x2=(b+x1)/2
x3 = (x2+x1)/2...

Ok, so I did the same thing as above until x5 and substituted all that are needed to be substituted with a and b.And here's what I get:

(a+b)/2,(a+3b)/4,(3a+5b)/8,(5a+11b)/16,(11a+21b)/32+...

I realize a pattern here. Obviously, the denominator is just power of 2. I see a pattern in the numerator too,but it's hard to generalize it using n.I realize that the value of the second a is equals to the value of the previous b and the value of the third a is equals to the second b and it goes on.
You see it too,don't you?When the first term has b in the numerator,then the second term has a, and when the 2nd term has 3b in the numerator, the 3rd term has 3a and so on...now we just have to generalize that,don't we?
 
Last edited:
Physics news on Phys.org
x_n = \gamma_n(\alpha_n a + \beta_n b)

you have figured out that \gamma_n = 2^{-n}

You've noticed that \alpha_n = \beta_{n-1}

The pattern of the coefficients is 0, 1, 1, 3, 5, 11, 21, ... looks kinda familiar doesn't it?
Almost a Fibonacci series but actually it's the Jacobsthal sequence.

J_n = J_{n-1} + 2 J_{n-2} \; : \; J_0 = 0, J_1 = 1
 
Last edited:
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top