Geometric Sequences And Series

I.  Homework

II.  Geometric Sequences

Exercise: 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, ...

Definition:  A geometric series is a series with a common quotient, r, ie., a_n = ra_n-1  

Theorem:  

The general element of a geometric series is

a_n = a₁r^n-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.  Hence

a_n = a₁2^{n - 1}

Hence 80 = a₁2⁴ so that a₁ = 5

We see that  

a_n = 5(2^{n - 1})

Finally a₃ = 5(2²) = 40

The Geometric Series:      

Theorem:  sum from i = 1 to n of a₁r^i-1  = a₁(1 - rⁿ)/(1 - r)  

We proved this last week.

Example:  Find sum from i = 1 to 20 of 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)

Application:  Annuities:

I pay $300 per month into an annuity till I am 65 years old.  The annuity earns 6% interest.  How much will I get at age 65?

Solution:

We consider the process backwards.  The last months 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)⁴⁰⁸    

= sum from i = 1 to 409 of  300(1.005)ⁱ = 300[(1 - 1.005⁴⁰⁹)/(1 - 1.005)]

= $401,392