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**>> 10.4. Mathematical Induction**** **

10.3. Geometric Sequences

Another important type of sequence are geometric sequences. A * geometric sequence* is a sequence of the form a

_{0}= a, a

_{1}= ar, a

_{2}= ar

^{2}, a

_{3}= ar

^{3}, ..., where a and r are nonzero constants. We see that the term with index k is a

_{k}= ar

^{k}. Thus, like with arithmetic sequences, it is easy to compute terms of a geometric sequence with known index if we know a and r. But a and r are easy to compute. Assuming the sequence begins with index 0, a is the first term of the sequence. Also, r is the ratio of consecutive terms in the sequence (r = a

_{k+1}/ a

_{k}).

**Example 1:** For each of the following geometric sequences, compute a_{5} and a_{10}. (Assume each sequence starts with index 0.)

- (a) 1, 2, 4, 8, ...
- (b) 1, 10, 100, 1000, ...
- (c) 1, 1/2, 1/4, 1/8, ...
- (d) 1, -1/2, 1/4, -1/8, ...
- (e) 3, 9, 27, 81, ...
- (f) 3, 6, 12, 24, ...

**Solution: **

- (a) For this sequence, we have a = 1 and r = 2. Thus, we have a
_{k}= 2^{k}, implying a_{5}= 2^{5}= 32 and a_{10}= 2^{10}= 1024. - (b) For this sequence, we have a = 1 and r = 10. Thus, we have a
_{k}= 10^{k}, implying a_{5}= 10^{5}= 100,000 and a_{10}= 10^{10}= 10,000,000,000. - (c) For this sequence, we have a = 1 and r = 1/2. Thus, we have a
_{k}= (1/2)^{k}= 1/2^{k}, implying a_{5}= 1/2^{5}= 1/32 and a_{10}= 1/2^{10}= 1 / 1024. - (d) For this sequence, we have a = 1 and r = -1/2. Thus, we have a
_{k}= (-1/2)^{k}= (-1)^{k}/ 2^{k}, implying a_{5}= (-1)^{5}/ 2^{5}= -2^{5}= -1/32 and a_{10}= (-1)^{10 }/ 2^{10}= 1 / 1024. - (e) For this sequence, we have a = r = 3. Thus, we have a
_{k}= 3 * 3^{k}= 3^{k+1}, implying a_{5}= 3^{6}= 729 and a_{10}= 3^{11}= 177,147. - (f) For this sequence, we have a = 3 and r = 2. Thus, we have a
_{k}= 3 * 2^{k}, implying a_{5}= 3 * 2^{5}= 96 and a_{10}= 3 * 2^{10}= 3072.

As is the case with arithmetic sequences, there is a formula for computing series of geometric sequences, known as * geometric series*. The formula is as follows. Here we assume a and r are arbitrary real numbers with r≠1.

- (10.3.1)

It is fairly easy to derive Equation (10.3.1). Consider the geometric series above; call it S. Writing out all the terms, we have

S = a + ar + ar^{2} + ar^{3} + ... + ar^{n}.

Multiplying this series by r, we find

rS = ar + ar^{2} + ar^{3} + ... + ar^{n+1}.

Now subtracting the first equation from the second, we find

(r - 1)S = ar + ar^{2} + ar^{3} + ... + ar^{n+1} - a - ar - ar^{2} + ar^{3} - ... - ar^{n} = ar^{n+1} - a = a(r^{n+1} - 1).

Note that all the middle terms cancel, leaving behind only the last term of rS minus the first term of S. Dividing both sides by r - 1 leads to (10.3.1).

**Example 2: **Use Equation (10.3.1) to compute the sum of the terms of each sequence in Example 1 with index up to 10, i.e. the first 11 terms of each sequence.

**Solution: **

- (a) We have S = (1)(2
^{11}- 1) / (2 - 1) = 2^{11}- 1 = 2048 - 1 = 2047. - (b) We have S = (1)(10
^{11}- 1) / (10 - 1) = (10^{11}- 1) / 9 = 99,999,999,999 / 9 = 11,111,111,111. - (c) We have S = (1)[(1/2)
^{11}- 1] / (1/2 - 1) = (1 - 1/2^{11}) / (1/2) = (2^{11}- 1) / 2^{10}= 2047 / 1024. - (d) We have S = (1)[(-1/2)
^{11}- 1] / (-1/2 - 1) = (1 + 1/2^{11}) / (3/2) = (2^{11}+ 1) / (3 * 2^{10}) = 2049 / 3072 = 683 / 1024. - (e) We have S = (3)(3
^{11}- 1) / (3 - 1) = (3)(3^{11}- 1) / 2= (3)(177,146) / 2 = 265,719. - (f) We have S = (3)(2
^{11}- 1) / (2 - 1) = (3)(2^{11}- 1) = (3)(2047) = 6141.

It is instructive to consider the case of geometric series in which |r| < 1. For these sequences, it is better to write Equation (10.3.1) as follows:

Note that as n gets large, r^{n+1} gets smaller and smaller, approaching zero as n goes to infinity. Thus, in the * limit* in which n goes to infinity, the series can be expressed as follows:

- (10.3.2)

Thus we see that if |a| is less than 1, the infinite geometric series has a finite value. Such geometric series are said to * converge*.

**Example 3:** Use Equation (10.3.2) to compute the following infinite series:

- (a) S = 1 + 1/2 + 1/4 + 1/8 + ...
- (b) S = 3/10 + 3/100 + 3/1000 + ...
- (c) S = 1 - 1/2 + 1/4 - 1/8 +- ...
- (d) S = 2/3 + 4/9 + 8/27 + ...
^{}

**Solution: **

- (a) For this series, we have a = 1 and r = 1/2, whence S = 1/(1 - 1/2) = 1/(1/2) = 2.
- (b) For this series, we have a = 3/10 and r = 1/10, whence S = (3/10) / (1 - 1/10) = (3/10) / (9/10) = 3/9 = 1/3.
- (c) For this series, we have a = 1 and r = -1/2, whence S = 1/(1 + 1/2) = 1/(3/2) = 2/3.
- (c) For this series, we have a = r = 2/3, whence S = (2/3) / (1 - 2/3) = (2/3) / (1/3) = 2.

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