Given a sequence x_n, how do you find x_(n+1)? For example, I have two sequences a_n = ((1*3*5*...*(2n-1))/(2*4*6*...*(2n)) and b_n = ((2n+1)!/10^n), and respectively a_n+1 = ((1*3*5*...*(2n-1)(2n+1))/(2*4*6*...*(2n)(2n+2)) and b_n+1 = ((2n+1)!(2n+3)/(10^n)*10). I don't seem to notice what is the pattern. Could someone explain it?

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A sequence is represented by a0, a1, a2​​​​​​​, a3​​​​​, ..., an, where an=f(n) is a possible function for determining an generally. But this is only one of a number of ways of defining an. For example, a Fibonacci series defines an=an-2+an-1 and an cannot be defined (exactly) by using a function of n, although there is a formula for an approximation of an.

But if an is given as a function of n, then an+1 can easily be found by substitution.

We​​​​​ can also find an+1-an and thereby establish a relationship between consecutive terms. A simple example is, given an=(2n+1)/(2n-1), the series would be:

-1, 3, 5/3, 7/5, 9/7, ...

an+1=(2(n+1)+1)/(2(n+1)-1)=(2n+3)/(2n+1);

an+1-an=(2n+3)/(2n+1)-(2n+1)/(2n-1)=[(2n+3)(2n-1)-(2n+1)2]/(4n2-1),

an+1-an=(4n2+4n-3-4n2-4n-1)/(4n2-1)=-4/(4n2-1), so:

an+1=an-4/(4n2-1).

Now take one of your examples. an=(1)(3)(5)...(2n+1)/[(2)(4)(6)...(2n)]. 

an+1=(1)(3)(5)...(2n+3)/[(2)(4)(6)...(2n+2)].

an+1-an=(1)(3)(5)...(2n+3)/[(2)(4)(6)...(2n+2)]-(1)(3)(5)...(2n+1)/[(2)(4)(6)...(2n)],

an+1=an+[(2n+3)/(2n+2)]{(1)(3)(5)...(2n+1)/[(2)(4)(6)...(2n)]}. Although it's complicated, there is a clear function of n which links successive terms of the series. This series could also be expressed:

Given a0=1, an+1=(2n+3)/(2n+2)an, generating the series an=1, 3/2, 15/8, 105/48=35/16, ... Note that this is different from an+1=an+f(n), which is a sum, not a product.

I think the problem you have observed is, not how to calculate xn+1, but how to calculate an as f(n), a function of n. If and when you can do so, then you will be able to calculate another function of n, g(n), which relates successive terms in the series.

by Top Rated User (1.2m points)

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