Geometric Sequences And Series

1. Geometric Sequences

   Example: Find the General Element
   
       i)  Recursively          ii)  Explicitly

           A)  3, 6, 12, 24, 48, ...          B)  5, 15, 45, 135, ...
   
           C)  -3, 30, -300, 3000, ...     D)  2, 2/3, 2/9, 2/27, ...

   
   Solution

    

   1. We see that to get to the next term, we only need to multiply the previous term by 2, hence
      
              a_n = 2a_n-1,   a₁ = 3
      
      To find the n^th term, we see that 3 is multiplied by two n - 1 times.  This is the same as multiplying by 2^n-1.  Hence,
      
              a_n = 3(2^n-1)
      
      

   2. We see that to get to the next term, we only need to multiply the previous term by 3, hence
      
              a_n = 3a_n-1,   a₁ = 5
      
      To find the n^th term, we see that 5 is multiplied by three n - 1 times.  This is the same as multiplying by 3^n-1.  Hence,
      
              a_n = 5(3^n-1)
      
      

   3. This sequence follows a similar pattern with a₁ = -3 and a common ratio of -10.  Recursively we write
      
            a_n = -10a_n-1,   a₁ = -3
      
      and explicitly
      
               a_n = -3[(-10)^n-1]
      
      

   4. This sequence follows a similar pattern with a₁ = 2 and a common ratio of 1/3.  Recursively we write
      
            a_n = 1/3 a_n-1,   a₁ = 2
      
      and explicitly
      
               a_n = 2[(1/3)^n-1]

   
   

   Definition:   A geometric sequence is a sequence with a common quotient, r, ie.,            an = ran-1

    

   
   

   Theorem:   The general element of a geometric sequence is           an = a1rn-1

    

   
   Proof:  
   
   (by induction)
   For n = 1 it is trivial
   Assume that the theorem is true for n = k - 1 then 
   
           a_{k - 1} = a₁r^k-2
   
   Our goal is to show that
   
           a_k = a₁r^k-1
   
   but
   
           a_k = ra_k-1 = r(a₁r^k-2) = a₁r^k-1
   
   Hence by induction the theorem is true.
   

    

   Example:  
   
   Suppose that the fifth term of a geometric sequence is 80 and the eleventh term is 5120.  Find the third term
   

   Solution:  
   
   We have that 
   
           80 = a₁r⁴  
   
   and 
   
           5120 =  a₁r¹⁰  

   Dividing the equation, we get:  
   
           5120/80 = r⁶  
   
   or 
   
           r⁶ = 64 
   
   taking sixth roots we get 
   
           r = 2     or     r = -2
   
    Hence
   
           a_n = a₁2^{n - 1}      or    a_n = a₁(-2)^{n - 1}
   
   From this
   
           80 = a₁2⁴  
   
   so that 
   
           a₁ = 5
   
   We see that  
   
           a_n = 5(2^{n - 1})     or      a_n = 5((-2)^{n - 1})
   
   Finally 
   
           a₃ = 5(2²) = 20    (note that 5((-2)²) = 20 also)

   
   

2. The Geometric Series:      

   
   

   Theorem      S i = 1n  a1ri-1  = a1(1 - rn)/(1 - r)

    

   Proof:   
   
   By induction
   
   For n = 1, 
   
           a₁ = a₁(1 - r¹)/(1 - r) = a₁ 
   
   Now assume the theorem is true for n = k - 1, then
   
           S _{i = 1}^k-1  a₁r^i-1  = a₁(1 - r^k-1)/(1 - r)
   
   Our goal is to show that 
   
           S _{i = 1}^k  a₁r^i-1  = a₁(1 - r^k)/(1 - r)
   
   The left hand side is 
   
           S _{i = 1}^k  a₁r^i-1  = (S _{i = 1}^k-1  a₁r^i-1) + a₁r^k-1 
   
           = a₁(1 - r^k-1)/(1 - r) + a₁r^k-1 
   
           = [a₁/(1 - r)](1 - r^k-1 + (1 - r)r^k-1)
   
           = [a₁/(1 - r)](1 - r^k-1 + r^k-1 - r^k)
   
           = [a₁/(1 - r)](1 - r^k) = a₁(1 - r^k)/(1 - r)
   
   By induction the theorem is true.

    

   Example:  
   
   Find 
   
           S _{i = 1}²⁰ [3(1/2)ⁱ]

    

   Solution:

   We see that 
   
           a₁ = 3     and     r = 1/2
   
   Hence the sum is 
   
           3(1 - (1/2)²⁰)/(1 - 1/2) @ 5.999994

   
   

3. Application:  Annuities
   
   I pay $300 per month into an annuity till I am 65 years old (a total of 34 years).  The annuity earns 6% interest. How much will I get at age 65?
   

   Solution:
   
   We consider the process backwards.  The last month's payment will accrue no interest, the second to last month's payment will accrue 2 months of interest, the third to last month's payment will accrue 3 month's of interest, and so on.  Putting these all into the compound interest formula and adding them we get

           300 + 300(1 + .06/12) + 300(1 + .06/12)²   
               + 300(1 + .06/12)³ + ... + 300(1 + .06/12)⁴⁰⁷  (notice that there are (34)(12) = 408 payments and 0, ..., 407 are 408 numbers)
   
           = S_i=1⁴⁰⁸[300(1.005)^i-1] = 300[(1 - 1.005⁴⁰⁸)/(1 - 1.005)]
   
           = $399,096.99  

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