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Sequence Arithmetic progression [closed]

-1

$begingroup$

A sequence of positive number defined by $$begincasesa_n+1&=sqrta_n^2+3quad text for n≥1\a_1&=1endcases$$

• (a) Find $a_2$ and $a_3$ leaving your answer in surd form




• (b) Show that $a_5=4$

sequences-and-series

share|cite|improve this question

edited Mar 18 at 12:04

MarianD

1,7391617

asked Mar 18 at 10:54

RicknathanRicknathan

2

$endgroup$

closed as off-topic by Saad, trancelocation, StubbornAtom, Martin R, John Omielan Mar 18 at 12:30

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, trancelocation, StubbornAtom, Martin R, John Omielan

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Try putting $n=1$ into the recursive formula to find $a_2$.
  $endgroup$
  – Minus One-Twelfth
  Mar 18 at 10:56
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Is there a particular difficulty you are facing in this problem?
  $endgroup$
  – Matti P.
  Mar 18 at 10:57
  

  
  
  
  

add a comment | 

-1

$begingroup$

A sequence of positive number defined by $$begincasesa_n+1&=sqrta_n^2+3quad text for n≥1\a_1&=1endcases$$

• (a) Find $a_2$ and $a_3$ leaving your answer in surd form




• (b) Show that $a_5=4$

sequences-and-series

share|cite|improve this question

edited Mar 18 at 12:04

MarianD

1,7391617

asked Mar 18 at 10:54

RicknathanRicknathan

2

$endgroup$

closed as off-topic by Saad, trancelocation, StubbornAtom, Martin R, John Omielan Mar 18 at 12:30

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, trancelocation, StubbornAtom, Martin R, John Omielan

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Try putting $n=1$ into the recursive formula to find $a_2$.
  $endgroup$
  – Minus One-Twelfth
  Mar 18 at 10:56
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Is there a particular difficulty you are facing in this problem?
  $endgroup$
  – Matti P.
  Mar 18 at 10:57
  

  
  
  
  

add a comment | 

$begingroup$

A sequence of positive number defined by $$begincasesa_n+1&=sqrta_n^2+3quad text for n≥1\a_1&=1endcases$$

• (a) Find $a_2$ and $a_3$ leaving your answer in surd form




• (b) Show that $a_5=4$

sequences-and-series

share|cite|improve this question

edited Mar 18 at 12:04

MarianD

1,7391617

asked Mar 18 at 10:54

RicknathanRicknathan

2

$endgroup$

share|cite|improve this question

edited Mar 18 at 12:04

MarianD

1,7391617

asked Mar 18 at 10:54

RicknathanRicknathan

2

share|cite|improve this question

edited Mar 18 at 12:04

MarianD

1,7391617

asked Mar 18 at 10:54

RicknathanRicknathan

2

edited Mar 18 at 12:04

MarianD

1,7391617

edited Mar 18 at 12:04

MarianD

1,7391617

asked Mar 18 at 10:54

RicknathanRicknathan

2

asked Mar 18 at 10:54

RicknathanRicknathan

2

2 Answers
2

active

oldest

votes

1

$begingroup$

"By hand",

$$a_1=1,a_2=sqrt4=2,a_3=sqrt7,a_4=sqrt10,a_5=sqrt13,a_6=sqrt16=4.$$

There is a typo in your question.

share|cite|improve this answer

answered Mar 18 at 11:45

Yves DaoustYves Daoust

131k676229

$endgroup$

add a comment | 

0

$begingroup$

Hint

Define $$b_n=a_n^2quad,quad b_1=a_1^2=1$$then $$b_n+1=b_n+3$$is an arithmetic progression. Also $$a_5=sqrt13ne 4$$probably you mean $a_6$.

share|cite|improve this answer

answered Mar 18 at 11:07

Mostafa AyazMostafa Ayaz

18.1k31040

$endgroup$

add a comment | 

1

$begingroup$

"By hand",

$$a_1=1,a_2=sqrt4=2,a_3=sqrt7,a_4=sqrt10,a_5=sqrt13,a_6=sqrt16=4.$$

There is a typo in your question.

share|cite|improve this answer

answered Mar 18 at 11:45

Yves DaoustYves Daoust

131k676229

$endgroup$

add a comment | 

1

$begingroup$

"By hand",

$$a_1=1,a_2=sqrt4=2,a_3=sqrt7,a_4=sqrt10,a_5=sqrt13,a_6=sqrt16=4.$$

There is a typo in your question.

share|cite|improve this answer

answered Mar 18 at 11:45

Yves DaoustYves Daoust

131k676229

$endgroup$

add a comment | 

$begingroup$

"By hand",

$$a_1=1,a_2=sqrt4=2,a_3=sqrt7,a_4=sqrt10,a_5=sqrt13,a_6=sqrt16=4.$$

There is a typo in your question.

share|cite|improve this answer

answered Mar 18 at 11:45

Yves DaoustYves Daoust

131k676229

$endgroup$

share|cite|improve this answer

answered Mar 18 at 11:45

Yves DaoustYves Daoust

131k676229

answered Mar 18 at 11:45

Yves DaoustYves Daoust

131k676229

answered Mar 18 at 11:45

Yves DaoustYves Daoust

131k676229

0

$begingroup$

Hint

Define $$b_n=a_n^2quad,quad b_1=a_1^2=1$$then $$b_n+1=b_n+3$$is an arithmetic progression. Also $$a_5=sqrt13ne 4$$probably you mean $a_6$.

share|cite|improve this answer

answered Mar 18 at 11:07

Mostafa AyazMostafa Ayaz

18.1k31040

$endgroup$

add a comment | 

0

$begingroup$

Hint

Define $$b_n=a_n^2quad,quad b_1=a_1^2=1$$then $$b_n+1=b_n+3$$is an arithmetic progression. Also $$a_5=sqrt13ne 4$$probably you mean $a_6$.

share|cite|improve this answer

answered Mar 18 at 11:07

Mostafa AyazMostafa Ayaz

18.1k31040

$endgroup$

add a comment | 

$begingroup$

Hint

Define $$b_n=a_n^2quad,quad b_1=a_1^2=1$$then $$b_n+1=b_n+3$$is an arithmetic progression. Also $$a_5=sqrt13ne 4$$probably you mean $a_6$.

share|cite|improve this answer

answered Mar 18 at 11:07

Mostafa AyazMostafa Ayaz

18.1k31040

$endgroup$

share|cite|improve this answer

answered Mar 18 at 11:07

Mostafa AyazMostafa Ayaz

18.1k31040

answered Mar 18 at 11:07

Mostafa AyazMostafa Ayaz

18.1k31040

answered Mar 18 at 11:07

Mostafa AyazMostafa Ayaz

18.1k31040