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Summation of a Sequence

  • Thread starter S.R
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  • #1
S.R
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Homework Statement


What is the sum of:
2a2vF.png



Homework Equations


N/A


The Attempt at a Solution


I'm unsure how to start.

Note: I'm in Grade 10, so I may not have the mathematical skills necessary to understand the solutions you provide.

Any help/guidance would be appreciated.
 

Answers and Replies

  • #2
tiny-tim
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Hi S.R.! :smile:

Hint: each term is 1/√n√n+1(√n + √n+1) …

but what is 1/(√n + √n+1) ? :wink:
 
  • #3
S.R
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Hi S.R.! :smile:

Hint: each term is 1/√n√n+1(√n + √n+1) …

but what is 1/(√n + √n+1) ? :wink:
Assuming n+1 isn't inclusive: 1/(2√n+1). I'm not sure what to do with this information (if I'm correct, that is).
 
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  • #4
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Assuming n+1 isn't inclusive: 1/(2√n+1). I'm not sure what to do with this information (if I'm correct, that is).
Nope, n+1 is inclusive.

The general term is

[itex]T_n = \frac{1}{(\sqrt{n}\sqrt{n+1})(\sqrt{n} + \sqrt{n+1})}[/itex]

But, how can you simplify this part of the above equation?

[itex]\frac{1}{(\sqrt{n} + \sqrt{n+1})}[/itex]

Hint:Rationalize...
 
  • #5
tiny-tim
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another hint:

nobody likes square-roots on the bottom

nobody minds square-roots on the top :wink:
 
  • #6
S.R
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Nope, n+1 is inclusive.

The general term is

[itex]T_n = \frac{1}{(\sqrt{n}\sqrt{n+1})(\sqrt{n} + \sqrt{n+1})}[/itex]

But, how can you simplify this part of the above equation?

[itex]\frac{1}{(\sqrt{n} + \sqrt{n+1})}[/itex]

Hint:Rationalize...
Oh of course, √(n+1)-√n.
 
  • #7
881
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Oh of course, √(n+1)-√n.
Yep! Now, what did you do next?

Edit : maybe its just too simple from here :biggrin:
 
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  • #8
S.R
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Yep! Now, what did you do next?

Edit : maybe its just too simple from here :biggrin:
Simplifying the general expression, T(n)=(sqrt(n+1)-sqrt(n))^2. However, I'm stuck here. Is there an applicable formula?
 
  • #9
881
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Simplifying the general expression, T(n)=(sqrt(n+1)-sqrt(n))^2. However, I'm stuck here. Is there an applicable formula?
Uhh, where'd get that whole square from?? :confused:

The simplification from the general term will yield you a difference of two terms. You can write them as

[itex]T_1 = A_2 - A_1[/itex]
[itex]T_2 = A_3 - A_2[/itex]

and so on. The sum of all terms is the sum of the series. Can you notice something in the above equations that makes solving this easier?
 
  • #10
tiny-tim
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Simplifying the general expression, T(n)=(sqrt(n+1)-sqrt(n))^2.
nooo, Tn = (√(n+1) - √n)/√n√(n+1) = … ? :smile:
 
  • #11
S.R
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Sorry I misread from my iPhone. However, the simplification from the general term is sqrt(n+1)-sqrt(n)/sqrt(n)sqrt(n+1) = sqrt(n)-sqrt(n+1).

From your explanation I noticed, Tn=An+1-An.

Note that the extra terms are suppose to be subscrippts.
 
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  • #12
881
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Sorry I misread from my iPhone. However, the simplification from the general term is sqrt(n+1)-sqrt(n)/sqrt(n)sqrt(n+1) = sqrt(n)-sqrt(n+1).

From your explanation I noticed, Tn=Tn+1-Tn.

Note that the extra terms are suppose to be subscrippts.
That would make [itex]2T_n = T_{n+1}[/itex] which is untrue. :wink: Its better to use different term letters for it. So, what do you get from that relation, by summing it all up?
 
  • #13
tiny-tim
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Sorry I misread from my iPhone. However, the simplification from the general term is sqrt(n+1)-sqrt(n)/sqrt(n)sqrt(n+1) = sqrt(n)-sqrt(n+1).
no!!!!!
 
  • #14
S.R
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no!!!!!
Im unsure why not? Im not sure if my division is correct though.

Edit: Sorry, I was replying in English class and got distracted :smile:.
 
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  • #15
881
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Im unsure why not? Im not sure if my division is correct though.
Its wrong :tongue2:

[tex]Tn = \frac{\sqrt{n+1} - \sqrt{n}}{\sqrt{n}\sqrt{n+1}}[/tex]


Give it another go, and write out the answer :smile:
 
  • #16
S.R
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Its wrong :tongue2:

[tex]Tn = \frac{\sqrt{n+1} - \sqrt{n}}{\sqrt{n}\sqrt{n+1}}[/tex]


Give it another go, and write out the answer :smile:
Tn=1/sqrt(n)-1/sqrt(n+1)?
 
  • #17
881
40
Tn=1/sqrt(n)-1/sqrt(n+1)?
Yep. Now apply the logic I suggested in post #9 and #12.
 
  • #18
S.R
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Yep. Now apply the logic I suggested in post #9 and #12.
I don't notice any patterns to find the sum?
 
  • #19
22,097
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I don't notice any patterns to find the sum?
Can you write out the first ten terms of the sum to see if you notice anything??
 
  • #20
S.R
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Can you write out the first ten terms of the sum to see if you notice anything??
The sum is 9/10. Correct?
 
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  • #21
S.R
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I still don't understand how you came up with the general form of the sequence, though?
 
  • #22
tiny-tim
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I still don't understand how you came up with the general form of the sequence, though?
what is [tex]\Sigma_5^{21}\ \left(\frac{1}{n}-\frac{1}{n+1}\right)[/tex] ? :wink:
 
  • #23
S.R
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what is [tex]\Sigma_5^{21}\ \left(\frac{1}{n}-\frac{1}{n+1}\right)[/tex] ? :wink:
The terms 1/5 and -1/24 are left after summation, therefore 19/120. However, my question is how did you obtain: Tn=1/(sqrt(n)sqrt(n+1))((sqrt(n+1)+sqrt(n))? Sorry for the notation.
 
  • #24
tiny-tim
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The terms 1/5 and -1/24 are left after summation, therefore 19/120. However, my question is how did you obtain: Tn=1/(sqrt(n)sqrt(n+1))((sqrt(n+1)+sqrt(n))? Sorry for the notation.
√n√(n+1)(√n + √(n+1)) = n√(n+1) + (n+1)√n
 
  • #25
S.R
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√n√(n+1)(√n + √(n+1)) = n√(n+1) + (n+1)√n
Thanks!
 

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