Geometric Sequences And Series

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
We see that to get to the next term, we only need to
multiply the previous term by 2, hence
a_{n} = 2a_{n1},
a_{1} = 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^{n1}.
Hence,
a_{n} = 3(2^{n1})
We see that to get to the next term, we only need to
multiply the previous term by 3, hence
a_{n} = 3a_{n1},
a_{1} = 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^{n1}.
Hence,
a_{n} = 5(3^{n1})
This sequence follows a similar pattern with a_{1}
= 3 and a common ratio of 10. Recursively we write
a_{n} = 10a_{n1},
a_{1} = 3
and explicitly
a_{n} =
3[(10)^{n1}]
This sequence follows a similar pattern with a_{1}
= 2 and a common ratio of 1/3. Recursively we write
a_{n} = 1/3 a_{n1},
a_{1} = 2
and explicitly
a_{n} = 2[(1/3)^{n1}]
Definition:
A geometric sequence is a
sequence with a common quotient,
r, ie.,
a_{n} = ra_{n1} 
Theorem:
The general element of a geometric sequence is
a_{n} = a_{1}r^{n1} 
Proof:
(by induction)
For n = 1 it is trivial
Assume that the theorem is true for n = k  1 then
a_{k  1} = a_{1}r^{k2
}Our goal is to show that
a_{k} = a_{1}r^{k1
}but
a_{k} = ra_{k1} = r(a_{1}r^{k2}) =
a_{1}r^{k1
}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 = a_{1}r^{4}
and
5120 = a_{1}r^{10}
Dividing the equation, we get:
5120/80 = r^{6}
or
r^{6} = 64
taking sixth roots we get
r = 2
or r
= 2
Hence
a_{n} = a_{1}2^{n  1}
or a_{n} = a_{1}(2)^{n  1
}From this
80 = a_{1}2^{4}
so that
a_{1} = 5
We see that
a_{n} = 5(2^{n  1})
or a_{n} =
5((2)^{n  1})
Finally
a_{3} = 5(2^{2}) = 20
(note that 5((2)^{2}) = 20 also)

The Geometric Series:
Theorem
S _{i = 1}^{n}
a_{1}r^{i1}
= a_{1}(1  r^{n})/(1  r) 
Proof:
By induction
For n = 1,
a_{1} = a_{1}(1 
r^{1})/(1  r) = a_{1}
Now assume the theorem is true for n = k  1, then
S _{ i = 1}^{k1}
a_{1}r^{i1}
= a_{1}(1  r^{k1})/(1  r)
Our goal is to show that
S _{ i = 1}^{k}
a_{1}r^{i1}
= a_{1}(1  r^{k})/(1  r)
The left hand side is
S _{ i = 1}^{k} a_{1}r^{i1}
= (S _{ i = 1}^{k1} a_{1}r^{i1}) +
a_{1}r^{k1}
= a_{1}(1  r^{k1})/(1  r)
+ a_{1}r^{k1}
= [a_{1}/(1  r)](1  r^{k1}
+ (1  r)r^{k1})
= [a_{1}/(1  r)](1  r^{k1}
+ r^{k1}  r^{k})
= [a_{1}/(1  r)](1  r^{k})
= a_{1}(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
a_{1} = 3
and r = 1/2
Hence the sum is
3(1  (1/2)^{20})/(1  1/2) @
5.999994

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=1}^{408}[300(1.005)^{i1}]
= 300[(1 
1.005^{408})/(1  1.005)]
= $399,096.99
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