How do I solve the summation of a sequence for my homework?

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Homework Help Overview

The discussion revolves around finding the sum of a sequence involving terms of the form \( T_n = \frac{1}{\sqrt{n}\sqrt{n+1}(\sqrt{n} + \sqrt{n+1})} \). Participants express uncertainty about their mathematical skills and seek guidance on how to approach the problem.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants attempt to simplify the general term and explore rationalization techniques. They question the correctness of their simplifications and seek clarification on the general form of the sequence.

Discussion Status

There is ongoing exploration of different simplifications and terms. Some participants have provided hints and guidance on rationalizing terms, while others are questioning their own understanding and the implications of their findings. Multiple interpretations of the sequence are being discussed.

Contextual Notes

Participants note that they are in Grade 10, which may influence their confidence in handling the mathematical concepts involved. There is also a mention of specific summation limits and terms being inclusive or exclusive, which affects the interpretation of the sequence.

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.
 
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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:​
 
tiny-tim said:
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).
 
Last edited:
S.R said:
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...
 
another hint:

nobody likes square-roots on the bottom

nobody minds square-roots on the top :wink:​
 
Infinitum said:
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.
 
S.R said:
Oh of course, √(n+1)-√n.

Yep! Now, what did you do next?

Edit : maybe its just too simple from here :biggrin:
 
Last edited:
Infinitum said:
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?
 
S.R said:
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
S.R said:
Simplifying the general expression, T(n)=(sqrt(n+1)-sqrt(n))^2.

nooo, Tn = (√(n+1) - √n)/√n√(n+1) = … ? :smile:
 
  • #11
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
S.R said:
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
S.R said:
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
tiny-tim said:
no!

Im unsure why not? I am 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
S.R said:
Im unsure why not? I am not sure if my division is correct though.

Its wrong :-p

[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
Infinitum said:
Its wrong :-p

[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
S.R said:
Tn=1/sqrt(n)-1/sqrt(n+1)?

Yep. Now apply the logic I suggested in post #9 and #12.
 
  • #18
Infinitum said:
Yep. Now apply the logic I suggested in post #9 and #12.
I don't notice any patterns to find the sum?
 
  • #19
S.R said:
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
micromass said:
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
I still don't understand how you came up with the general form of the sequence, though?
 
  • #22
S.R said:
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
tiny-tim said:
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
S.R said:
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
tiny-tim said:
√n√(n+1)(√n + √(n+1)) = n√(n+1) + (n+1)√n

Thanks!
 

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