Difference between revisions of "Geometric sequence"

(Infinite Geometric Sequences)
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==Definition==
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A '''geometric sequence''' is a [[sequence]] of numbers in which each term is a fixed [[multiple]] of the previous term.  For example: 1, 2, 4, 8, 16, 32, ...  is a geometric sequence because each term is twice the previous term.  In this case, 2 is called the ''common ratio'' of the sequence.  More formally, a geometric sequence may be defined [[recursion|recursively]] by:
  
A geometric sequence is a sequence of numbers in which each term is a fixed multiple of the previous term.  For example: 1, 2, 4, 8, 16, 32, ...  is a geometric sequence because each term is twice the previous term.  In this case, 2 is called the common ratio of the sequence.  More formally, a geometric sequence may be defined [[recursion|recursively]] by:
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<center><math>a_n = r\cdot a_{n-1}, n > 1</math></center>
  
<math>a_n = r\cdot a_{n-1}, n \geq 1</math>
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with a fixed first term <math>a_1</math> and common ratio <math>r</math>.  Using this definition, the <math>n</math>th term has the closed-form:
  
with a fixed <math>a_0</math> and common ratio <math>r</math>.  Using this definition, the <math>n</math>th term has the closed-form:
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<center><math>a_n = a_1\cdot r^{n-1}</math></center>
 
 
<math>\displaystyle a_n = a_0\cdot r^n</math>
 
  
 
==Summing a Geometric Sequence==
 
==Summing a Geometric Sequence==
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The sum of the first <math>n</math> terms of a geometric sequence is given by
 
The sum of the first <math>n</math> terms of a geometric sequence is given by
  
<math>S_n = a_0 + a_1 + \ldots + a_{n - 1} = a_0\cdot\frac{r^n-1}{r-1}</math>
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<center><math>S_n = a_1 + a_2 + \cdots + a_n = a_1\cdot\frac{r^n-1}{r-1}</math></center>
  
where <math>a_0</math> is the first term in the sequence, and <math>r</math> is the common ratio.
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where <math>a_1</math> is the first term in the sequence, and <math>r</math> is the common ratio.
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===Proof===
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The geometric sequence can be rewritten as <math> a_1+r \cdot a_1+r^2 \cdot a_1+ \cdots + r^{n-1} \cdot a_1=a_1(1+r+r^2+ \cdots +r^{n-1})</math> where <math>n</math> is the amount of terms, <math>r</math> is the common ratio, and <math>a_1</math> is the first term. Multiplying in <math>(r-1)</math> yields <math>r^n-1</math> so <math> a_1 + a_2 + \cdots + a_n = a_1\cdot\frac{r^n-1}{r-1} </math>.
  
 
==Infinite Geometric Sequences==
 
==Infinite Geometric Sequences==
  
An infinite geometric sequence is a geometric sequence with an infinite number of terms.  If the common ratio is small, the sum of the terms will approach a fixed [[limit]].  In this case, "small" means <math>|r|<1</math>.  We say that the sum of the terms of this sequence is a [[convergent sum]].
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An [[infinite]] geometric sequence is a geometric sequence with an infinite number of terms.  If the common ratio is small, the terms will approach 0 and the sum of the terms will approach a fixed [[limit]].  In this case, "small" means <math>|r|<1</math>.  We say that the sum of the terms of this sequence is a [[convergent|convergent sum]].
  
 
For instance, the series <math>1 + \frac12 + \frac14 + \frac18 + \cdots</math>, sums to 2.  The general formula for the sum of such a sequence is:
 
For instance, the series <math>1 + \frac12 + \frac14 + \frac18 + \cdots</math>, sums to 2.  The general formula for the sum of such a sequence is:
  
<math>S = \frac{a_0}{1-r}</math>
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<center><math>S = \frac{a_1}{1-r}</math>.</center> <br><br>
  
Again, <math>a_0</math> is the first term in the sequence, and <math>r</math> is the common ratio.
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Where <math>a_1</math> is the first term in the sequence, and <math>r</math> is the common ratio.
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===Proof===
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Let the sequence be
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<center><math>S=a_1+a_1r+a_1r^2+a_1r^3+\cdots.</math></center>   
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Multiplying by <math>r</math> yields,
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<center><math>S \cdot r=a_1r+a_1r^2+a_1r^3+\cdots.</math></center>
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We subtract these two equations to obtain:
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<center><math> S-Sr=a_1.</math></center>
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There is only one term on the RHS because the rest of the terms cancel out after subtraction. Finally, we can factor and divide to get
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<center><math>S(1-r)=a_1</math></center>
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thus,
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<center><math>S=\frac{a_1}{1-r}.</math></center>
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This method of multiplying the sequence and subtracting equations, called telescoping, is a frequently used method to evaluate infinite sequences.  In fact, the same method can be used to calculate the sum of a finite geometric sequence (given above).
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===Common uses===
  
 
One common instance of summing infinite geometric sequences is the [[decimal expansion]] of most [[rational number]]s.  For instance, <math>0.33333\ldots = \frac 3{10} + \frac3{100} + \frac3{1000} + \frac3{10000} + \ldots</math> has first term <math>a_0 = \frac 3{10}</math> and common ratio <math>\frac1{10}</math>, so the infinite sum has value <math>S = \frac{\frac3{10}}{1-\frac1{10}} = \frac13</math>, just as we would have expected.
 
One common instance of summing infinite geometric sequences is the [[decimal expansion]] of most [[rational number]]s.  For instance, <math>0.33333\ldots = \frac 3{10} + \frac3{100} + \frac3{1000} + \frac3{10000} + \ldots</math> has first term <math>a_0 = \frac 3{10}</math> and common ratio <math>\frac1{10}</math>, so the infinite sum has value <math>S = \frac{\frac3{10}}{1-\frac1{10}} = \frac13</math>, just as we would have expected.
  
==See Also==
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== Problems ==
[[arithmetic sequence|Arithmetic Sequences]]
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=== Intermediate ===
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* [[2005_AIME_II_Problems/Problem_3 | 2005 AIME II Problem 3]]
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* [[2007 AIME II Problems/Problem 12 | 2007 AIME II Problem 12]]
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==See also==
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*[[Arithmetic sequence]]
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*[[Sequence]]
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*[[Series]]

Revision as of 11:50, 19 December 2018

A geometric sequence is a sequence of numbers in which each term is a fixed multiple of the previous term. For example: 1, 2, 4, 8, 16, 32, ... is a geometric sequence because each term is twice the previous term. In this case, 2 is called the common ratio of the sequence. More formally, a geometric sequence may be defined recursively by:

$a_n = r\cdot a_{n-1}, n > 1$

with a fixed first term $a_1$ and common ratio $r$. Using this definition, the $n$th term has the closed-form:

$a_n = a_1\cdot r^{n-1}$

Summing a Geometric Sequence

The sum of the first $n$ terms of a geometric sequence is given by

$S_n = a_1 + a_2 + \cdots + a_n = a_1\cdot\frac{r^n-1}{r-1}$

where $a_1$ is the first term in the sequence, and $r$ is the common ratio.

Proof

The geometric sequence can be rewritten as $a_1+r \cdot a_1+r^2 \cdot a_1+ \cdots + r^{n-1} \cdot a_1=a_1(1+r+r^2+ \cdots +r^{n-1})$ where $n$ is the amount of terms, $r$ is the common ratio, and $a_1$ is the first term. Multiplying in $(r-1)$ yields $r^n-1$ so $a_1 + a_2 + \cdots + a_n = a_1\cdot\frac{r^n-1}{r-1}$.

Infinite Geometric Sequences

An infinite geometric sequence is a geometric sequence with an infinite number of terms. If the common ratio is small, the terms will approach 0 and the sum of the terms will approach a fixed limit. In this case, "small" means $|r|<1$. We say that the sum of the terms of this sequence is a convergent sum.

For instance, the series $1 + \frac12 + \frac14 + \frac18 + \cdots$, sums to 2. The general formula for the sum of such a sequence is:

$S = \frac{a_1}{1-r}$.



Where $a_1$ is the first term in the sequence, and $r$ is the common ratio.

Proof

Let the sequence be

$S=a_1+a_1r+a_1r^2+a_1r^3+\cdots.$

Multiplying by $r$ yields,

$S \cdot r=a_1r+a_1r^2+a_1r^3+\cdots.$

We subtract these two equations to obtain:

$S-Sr=a_1.$

There is only one term on the RHS because the rest of the terms cancel out after subtraction. Finally, we can factor and divide to get

$S(1-r)=a_1$

thus,

$S=\frac{a_1}{1-r}.$

This method of multiplying the sequence and subtracting equations, called telescoping, is a frequently used method to evaluate infinite sequences. In fact, the same method can be used to calculate the sum of a finite geometric sequence (given above).

Common uses

One common instance of summing infinite geometric sequences is the decimal expansion of most rational numbers. For instance, $0.33333\ldots = \frac 3{10} + \frac3{100} + \frac3{1000} + \frac3{10000} + \ldots$ has first term $a_0 = \frac 3{10}$ and common ratio $\frac1{10}$, so the infinite sum has value $S = \frac{\frac3{10}}{1-\frac1{10}} = \frac13$, just as we would have expected.

Problems

Intermediate

See also