Expert: Paul Klarreich - 12/16/2006

Question
A sequence of integers is defined as follows asub1=1 asub2=1

and   asubn=.25asub(n-2)+.333asub(n-1) for n=3,4,5,6....

Let S= asub1+asub2+asub3+asub4+asub5+asub6+...

Find S.


If you cannot help me it will be alright but this problem is driving me insane.

Answer
Questioner:  Xia
Category:  Calculus
 
Subject:  sequence
Question:  A sequence of integers is defined as follows asub1=1 asub2=1

and   asubn=.25asub(n-2)+.333asub(n-1) for n=3,4,5,6....

Let S= asub1+asub2+asub3+asub4+asub5+asub6+...

Find S.

If you cannot help me it will be alright but this problem is driving me insane.
...........................................
Hi, Xia,

Well, I have a good psychiatrist, but maybe we won't need her.  I have the outlines of a solution for you, and I think you will be able to complete it.

I will compress the notation a bit, and I assume that by writing .333 you really mean 1/3.

So I write  a1, a2, etc. for your asub1, asub2, or
a[n-2] for  asub(n-2) if needed.  They all mean subscripts.

You have  a0 = 1,  a1 = 1,  and (you see that I changed the initial subscripts to 0 and 1.)

an = a[n-2]/4 + a[n-1]/3

WARNING: THE FOLLOWING DISCUSSION MAY CONTAIN FRACTIONS AND OTHER MATERIAL INAPPROPRIATE FOR CERTAIN COMPUTING SYSTEMS.  VIEW IT IN A FIXED-SIZE FONT, SUCH AS COURIER.


This is called a RECURRENCE.  (The famous Fibonacci sequence is an example of one.)  The general scheme is to make the assumption that:

a[k] = x^k, for some x.  Then you write:

a[k] = a[k-1]/3 + a[k-2]/4

(You had these the other way around in the original -- almost drove me crazier.)

x^k = x^(k-1)/3 + x^(k-2)/4

Delete x^(k-2):

x^2 = x/3 + 1/4

x^2 - x/3 - 1/4 = 0

12x^2 - 4x - 3 = 0
   4 +- sqrt(160)
x = --------------
        24
   4 +- 4 sqrt(10)
x = ---------------
        24

   1 +- sqrt(10)
x = ---------------  [S is sqrt(10) to save typing.]
        6

A general solution is of the form  

a[k] = b x1^k + c x2^k, where x1 and x2 are the two solutions.
(I will write  x1 as the solution with '+')

Now  a0 = 1  and  a1 = 1, as the first two terms, given:

a0 = b + c = 1
a1 = bx1 + cx2 = 1
 1 + S     1 - S
b ----- + c ------ = 1
   6         6

b(1 + S) + c(1 - S) = 6
b + Sb + c - Sc = 6,  and  b + c = 1
S(b - c) = 5
        5    5S     S
b - c = --- = --- = ---
        S    10     2

        S
b - c = ---
        2
b + c =  1
------------
            S
2b    = 1 + ---
            2

       2 + S
2b    = -------
         2

   2 + S
b = -----
     4

c = 1 - b
       2 + S
c = 1 - -----
         4
   2 - S
c = -----
     4
OK, we can write the general solution now:

       2 + S   1 + S       2 - S  1 - S
a[k] =  ------ [-----]^k  + ----- [-----]^k.  [FGT]
         4       6           4      6

Check:

       2 + S   1 + S     2 - S  1 - S
a[1] =  ------ [-----]  + ----- [-----]
         4       6         4      6

       2 + 3S + S^2     2 - 3S + S^2
a[1] =  --------------  + --------------
             24              24

       4  + 2S^2
a[1] =  -----------
           24

       4  + 2(10)
a[1] =  ---------- = 1, check.
           24

Now, finally, how about your series?  [Notation: A SEQUENCE is a list of

numbers.  A SERIES is a summation of a list of numbers.]

I think I am going to leave this part to you.  Here's what you will do:

1. You have the formula for the general term. [FGT above.]
2. The summation has this form:

SUM (bx1^k + cx2^k)

= b SUM x1^k + c SUM x2^k

3. Proceed as follows:  SUM x1^k  is a geometric series, for which there is a general formula:
             1
SUM x1^k = -------
          1 - x1
and likewise for SUM x2^k

That should do it, after some more messy algebra.  If it still drives you crazy, I'll send the phone number of my psychiatrist.