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Can someone help

2

2

$begingroup$

Consider the geometric sequence $1, a, a^2, a^3,dots$ Suppose that the sum of two consecutive terms in the sequence gives the next term in the sequence. Find $a$.

sequences-and-series

share|cite|improve this question

edited 2 hours ago

Lehs

7,07931664

asked 2 hours ago

lollollollol

221

New contributor

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$endgroup$

• 
  
  
  

  
  8
  

  
  
  
  
  
  

  
  $begingroup$
  So $1+a = a^2$ (make sure you know why!). Can you find $a$ from this?
  $endgroup$
  – Minus One-Twelfth
  2 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Son of Bonacci would know the answer right away.
  $endgroup$
  – dnqxt
  2 hours ago
  

  
  
  
  

add a comment | 

2

2

$begingroup$

Consider the geometric sequence $1, a, a^2, a^3,dots$ Suppose that the sum of two consecutive terms in the sequence gives the next term in the sequence. Find $a$.

sequences-and-series

share|cite|improve this question

edited 2 hours ago

Lehs

7,07931664

asked 2 hours ago

lollollollol

221

New contributor

lollol is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  8
  

  
  
  
  
  
  

  
  $begingroup$
  So $1+a = a^2$ (make sure you know why!). Can you find $a$ from this?
  $endgroup$
  – Minus One-Twelfth
  2 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Son of Bonacci would know the answer right away.
  $endgroup$
  – dnqxt
  2 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Consider the geometric sequence $1, a, a^2, a^3,dots$ Suppose that the sum of two consecutive terms in the sequence gives the next term in the sequence. Find $a$.

sequences-and-series

share|cite|improve this question

edited 2 hours ago

Lehs

7,07931664

asked 2 hours ago

lollollollol

221

New contributor

lollol is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited 2 hours ago

Lehs

7,07931664

asked 2 hours ago

lollollollol

221

New contributor

lollol is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited 2 hours ago

Lehs

7,07931664

asked 2 hours ago

lollollollol

221

New contributor

lollol is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited 2 hours ago

Lehs

7,07931664

edited 2 hours ago

Lehs

7,07931664

asked 2 hours ago

lollollollol

221

New contributor

lollol is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 2 hours ago

lollollollol

221

2 Answers
2

active

oldest

votes

3

$begingroup$

$a^n + a^n + 1 = a^n + 2; tag 1$

$a ne 0 Longrightarrow 1 + a = a^2 Longrightarrow a^2 - a - 1 = 0; tag 2$

$a_pm = dfrac1 pm sqrt(-1)^2 - 4(1)(-1)2 = dfrac1 pm sqrt 52; tag 3$

note that

$a_+ a_- = -1; ; a_+ + a_- = 1. tag 4$

share|cite|improve this answer

answered 1 hour ago

Robert LewisRobert Lewis

48.5k23167

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  In fact, $a=0$ is also a solution.
  $endgroup$
  – Ross Millikan
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @RossMillikan: except when we start at the beginning: $1 + 0 = 1 ne 0$!
  $endgroup$
  – Robert Lewis
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I read the question to have one set of three consecutive terms to satisfy the requirement, so the second, third, and fourth of $1,0,0,0,0,ldots$ do. I agree it is not clear and you might require that every set of three terms does.
  $endgroup$
  – Ross Millikan
  1 hour ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @RossMillikan: yes, I see your point!
  $endgroup$
  – Robert Lewis
  1 hour ago
  

  
  
  
  

add a comment | 

3

$begingroup$

$$1+a = a^2$$
$$textBy Quadratic formula, you get a = frac 1 pm sqrt52$$
You check that it works throughout the equation... as $a+a^2 = a^3$ and so on.. but you will find that $a^2 - a -1$ is always a factor of all equation.

Hence $a = frac 1 pm sqrt52$

share|cite|improve this answer

answered 1 hour ago

rashrash

595116

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The question asks if any set of three terms satisfies the requirement. If $a_n$ is the first one, you can divide by $a_n$ to get your equation, but that should be noted.
  $endgroup$
  – Ross Millikan
  1 hour ago
  

  
  
  
  

add a comment | 

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3

$begingroup$

$a^n + a^n + 1 = a^n + 2; tag 1$

$a ne 0 Longrightarrow 1 + a = a^2 Longrightarrow a^2 - a - 1 = 0; tag 2$

$a_pm = dfrac1 pm sqrt(-1)^2 - 4(1)(-1)2 = dfrac1 pm sqrt 52; tag 3$

note that

$a_+ a_- = -1; ; a_+ + a_- = 1. tag 4$

share|cite|improve this answer

answered 1 hour ago

Robert LewisRobert Lewis

48.5k23167

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  In fact, $a=0$ is also a solution.
  $endgroup$
  – Ross Millikan
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @RossMillikan: except when we start at the beginning: $1 + 0 = 1 ne 0$!
  $endgroup$
  – Robert Lewis
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I read the question to have one set of three consecutive terms to satisfy the requirement, so the second, third, and fourth of $1,0,0,0,0,ldots$ do. I agree it is not clear and you might require that every set of three terms does.
  $endgroup$
  – Ross Millikan
  1 hour ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @RossMillikan: yes, I see your point!
  $endgroup$
  – Robert Lewis
  1 hour ago
  

  
  
  
  

add a comment | 

3

$begingroup$

$a^n + a^n + 1 = a^n + 2; tag 1$

$a ne 0 Longrightarrow 1 + a = a^2 Longrightarrow a^2 - a - 1 = 0; tag 2$

$a_pm = dfrac1 pm sqrt(-1)^2 - 4(1)(-1)2 = dfrac1 pm sqrt 52; tag 3$

note that

$a_+ a_- = -1; ; a_+ + a_- = 1. tag 4$

share|cite|improve this answer

answered 1 hour ago

Robert LewisRobert Lewis

48.5k23167

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  In fact, $a=0$ is also a solution.
  $endgroup$
  – Ross Millikan
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @RossMillikan: except when we start at the beginning: $1 + 0 = 1 ne 0$!
  $endgroup$
  – Robert Lewis
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I read the question to have one set of three consecutive terms to satisfy the requirement, so the second, third, and fourth of $1,0,0,0,0,ldots$ do. I agree it is not clear and you might require that every set of three terms does.
  $endgroup$
  – Ross Millikan
  1 hour ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @RossMillikan: yes, I see your point!
  $endgroup$
  – Robert Lewis
  1 hour ago
  

  
  
  
  

add a comment | 

$begingroup$

$a^n + a^n + 1 = a^n + 2; tag 1$

$a ne 0 Longrightarrow 1 + a = a^2 Longrightarrow a^2 - a - 1 = 0; tag 2$

$a_pm = dfrac1 pm sqrt(-1)^2 - 4(1)(-1)2 = dfrac1 pm sqrt 52; tag 3$

note that

$a_+ a_- = -1; ; a_+ + a_- = 1. tag 4$

share|cite|improve this answer

answered 1 hour ago

Robert LewisRobert Lewis

48.5k23167

$endgroup$

share|cite|improve this answer

answered 1 hour ago

Robert LewisRobert Lewis

48.5k23167

answered 1 hour ago

Robert LewisRobert Lewis

48.5k23167

answered 1 hour ago

Robert LewisRobert Lewis

48.5k23167

3

$begingroup$

$$1+a = a^2$$
$$textBy Quadratic formula, you get a = frac 1 pm sqrt52$$
You check that it works throughout the equation... as $a+a^2 = a^3$ and so on.. but you will find that $a^2 - a -1$ is always a factor of all equation.

Hence $a = frac 1 pm sqrt52$

share|cite|improve this answer

answered 1 hour ago

rashrash

595116

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The question asks if any set of three terms satisfies the requirement. If $a_n$ is the first one, you can divide by $a_n$ to get your equation, but that should be noted.
  $endgroup$
  – Ross Millikan
  1 hour ago
  

  
  
  
  

add a comment | 

3

$begingroup$

$$1+a = a^2$$
$$textBy Quadratic formula, you get a = frac 1 pm sqrt52$$
You check that it works throughout the equation... as $a+a^2 = a^3$ and so on.. but you will find that $a^2 - a -1$ is always a factor of all equation.

Hence $a = frac 1 pm sqrt52$

share|cite|improve this answer

answered 1 hour ago

rashrash

595116

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The question asks if any set of three terms satisfies the requirement. If $a_n$ is the first one, you can divide by $a_n$ to get your equation, but that should be noted.
  $endgroup$
  – Ross Millikan
  1 hour ago
  

  
  
  
  

add a comment | 

$begingroup$

$$1+a = a^2$$
$$textBy Quadratic formula, you get a = frac 1 pm sqrt52$$
You check that it works throughout the equation... as $a+a^2 = a^3$ and so on.. but you will find that $a^2 - a -1$ is always a factor of all equation.

Hence $a = frac 1 pm sqrt52$

share|cite|improve this answer

answered 1 hour ago

rashrash

595116

$endgroup$

share|cite|improve this answer

answered 1 hour ago

rashrash

595116

answered 1 hour ago

rashrash

595116

answered 1 hour ago

rashrash

595116