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Represent a sequence with a Product sign

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0

$begingroup$

I have 2 sequences I have to represent with a Product symbol (Separately of course).

a) $2*sqrt3*sqrt4^3*sqrt5^4$

b) $dfrac1n*dfrac3n+2*dfrac7n+3$

I am clueless on what I should use as a start and end variable. $$prod_i=0^n i$$

For A I was thinking that it has something to do with 2, since you need 2 in the exponent to remove the square root.

Any suggestions?

sequences-and-series products

share|cite|improve this question

edited Mar 13 at 17:54

gt6989b

35k22557

asked Mar 13 at 17:48

ArioArio

184

$endgroup$

add a comment | 

0

$begingroup$

I have 2 sequences I have to represent with a Product symbol (Separately of course).

a) $2*sqrt3*sqrt4^3*sqrt5^4$

b) $dfrac1n*dfrac3n+2*dfrac7n+3$

I am clueless on what I should use as a start and end variable. $$prod_i=0^n i$$

For A I was thinking that it has something to do with 2, since you need 2 in the exponent to remove the square root.

Any suggestions?

sequences-and-series products

share|cite|improve this question

edited Mar 13 at 17:54

gt6989b

35k22557

asked Mar 13 at 17:48

ArioArio

184

$endgroup$

add a comment | 

$begingroup$

I have 2 sequences I have to represent with a Product symbol (Separately of course).

a) $2*sqrt3*sqrt4^3*sqrt5^4$

b) $dfrac1n*dfrac3n+2*dfrac7n+3$

I am clueless on what I should use as a start and end variable. $$prod_i=0^n i$$

For A I was thinking that it has something to do with 2, since you need 2 in the exponent to remove the square root.

Any suggestions?

sequences-and-series products

share|cite|improve this question

edited Mar 13 at 17:54

gt6989b

35k22557

asked Mar 13 at 17:48

ArioArio

184

$endgroup$

share|cite|improve this question

edited Mar 13 at 17:54

gt6989b

35k22557

asked Mar 13 at 17:48

ArioArio

184

share|cite|improve this question

edited Mar 13 at 17:54

gt6989b

35k22557

asked Mar 13 at 17:48

ArioArio

184

edited Mar 13 at 17:54

gt6989b

35k22557

edited Mar 13 at 17:54

gt6989b

35k22557

asked Mar 13 at 17:48

ArioArio

184

asked Mar 13 at 17:48

ArioArio

184

1 Answer
1

active

oldest

votes

1

$begingroup$

Products are like sums (and similar to integrals) in that the name of the index variable does not matter.

In your first sequence, the terms are $2^1=2^2/2,3^1/2,4^3/2,5^4/2$, I cannot obviously detect a pattern. But simpler one can serve as an example, say $2^1,3^1/2,4^1/3,5^1/4$, etc. It's easy to see the denominator in the exponent is one less than the base, and the base goes from 2 to 5. So you get
$$
prod_k=2^5 k^1/(k-1) = prod_k=1^4 (k+1)^1/k
$$

Can you now try both sequences after checking the first one for correctness?

share|cite|improve this answer

answered Mar 13 at 17:53

gt6989bgt6989b

35k22557

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Both solutions are correct when I check them. Thank you for the quick answer. Would you have any suggestions for the second sequence?
  $endgroup$
  – Ario
  Mar 14 at 13:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Ario I cannot make out the pattern for the second one
  $endgroup$
  – gt6989b
  Mar 14 at 14:20
  

  
  
  
  

add a comment | 

Your Answer

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1

$begingroup$

Products are like sums (and similar to integrals) in that the name of the index variable does not matter.

In your first sequence, the terms are $2^1=2^2/2,3^1/2,4^3/2,5^4/2$, I cannot obviously detect a pattern. But simpler one can serve as an example, say $2^1,3^1/2,4^1/3,5^1/4$, etc. It's easy to see the denominator in the exponent is one less than the base, and the base goes from 2 to 5. So you get
$$
prod_k=2^5 k^1/(k-1) = prod_k=1^4 (k+1)^1/k
$$

Can you now try both sequences after checking the first one for correctness?

share|cite|improve this answer

answered Mar 13 at 17:53

gt6989bgt6989b

35k22557

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Both solutions are correct when I check them. Thank you for the quick answer. Would you have any suggestions for the second sequence?
  $endgroup$
  – Ario
  Mar 14 at 13:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Ario I cannot make out the pattern for the second one
  $endgroup$
  – gt6989b
  Mar 14 at 14:20
  

  
  
  
  

add a comment | 

1

$begingroup$

Products are like sums (and similar to integrals) in that the name of the index variable does not matter.

In your first sequence, the terms are $2^1=2^2/2,3^1/2,4^3/2,5^4/2$, I cannot obviously detect a pattern. But simpler one can serve as an example, say $2^1,3^1/2,4^1/3,5^1/4$, etc. It's easy to see the denominator in the exponent is one less than the base, and the base goes from 2 to 5. So you get
$$
prod_k=2^5 k^1/(k-1) = prod_k=1^4 (k+1)^1/k
$$

Can you now try both sequences after checking the first one for correctness?

share|cite|improve this answer

answered Mar 13 at 17:53

gt6989bgt6989b

35k22557

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Both solutions are correct when I check them. Thank you for the quick answer. Would you have any suggestions for the second sequence?
  $endgroup$
  – Ario
  Mar 14 at 13:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Ario I cannot make out the pattern for the second one
  $endgroup$
  – gt6989b
  Mar 14 at 14:20
  

  
  
  
  

add a comment | 

$begingroup$

Products are like sums (and similar to integrals) in that the name of the index variable does not matter.

In your first sequence, the terms are $2^1=2^2/2,3^1/2,4^3/2,5^4/2$, I cannot obviously detect a pattern. But simpler one can serve as an example, say $2^1,3^1/2,4^1/3,5^1/4$, etc. It's easy to see the denominator in the exponent is one less than the base, and the base goes from 2 to 5. So you get
$$
prod_k=2^5 k^1/(k-1) = prod_k=1^4 (k+1)^1/k
$$

Can you now try both sequences after checking the first one for correctness?

share|cite|improve this answer

answered Mar 13 at 17:53

gt6989bgt6989b

35k22557

$endgroup$

share|cite|improve this answer

answered Mar 13 at 17:53

gt6989bgt6989b

35k22557

answered Mar 13 at 17:53

gt6989bgt6989b

35k22557

answered Mar 13 at 17:53

gt6989bgt6989b

35k22557