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

In summary, the sum of the given terms is 9/10. The general term is Tn = 1/(sqrt(n)sqrt(n+1))((sqrt(n+1)+sqrt(n)). This was obtained by rationalizing and simplifying the general term. The sum can also be written as ∑(1/n - 1/n+1) from n=5 to n=21.
  • #1
S.R
81
0

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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  • #2
Hi S.R.! :smile:

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

but what is 1/(√n + √n+1) ? :wink:​
 
  • #3
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).
 
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  • #4
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...
 
  • #5
another hint:

nobody likes square-roots on the bottom

nobody minds square-roots on the top :wink:​
 
  • #6
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.
 
  • #7
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:
 
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  • #8
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?
 
  • #9
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!
 

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

How do I approach solving a summation of a sequence?

The first step in solving a summation of a sequence is to identify the pattern or formula of the sequence. This can be done by examining the numbers in the sequence and looking for any relationships or repeating patterns. Once the pattern is identified, you can use it to generate the terms of the sequence and simplify the summation.

What are the different methods for solving a summation of a sequence?

There are several methods for solving a summation of a sequence, including the direct method, the difference method, and the substitution method. The direct method involves plugging in the values of each term in the sequence and adding them together. The difference method involves subtracting two consecutive terms to simplify the summation. The substitution method involves replacing the summation with an equivalent expression that is easier to solve.

How can I determine the upper and lower limits of a summation?

The upper and lower limits of a summation can be determined by looking at the given sequence and identifying the starting and ending values. The lower limit is usually the first term in the sequence, while the upper limit is the last term. If the sequence is given in terms of a variable, the limits can also be determined by the range of values for that variable.

Can I use a calculator to solve a summation of a sequence?

Yes, you can use a calculator to solve a summation of a sequence. However, it is important to know how to solve it manually as well. Some calculators have specific functions for summations, while others may require you to input the formula or pattern of the sequence. Additionally, make sure to check your calculator's settings to ensure that it is calculating the summation correctly.

What should I do if I am having trouble solving a summation of a sequence?

If you are having trouble solving a summation of a sequence, try breaking it down into smaller parts. Look for patterns in the sequence and try to simplify the summation using different methods. You can also ask for help from your teacher, classmates, or a tutor. Additionally, there are many online resources and tutorials available that can provide step-by-step guidance on how to solve different types of summations.

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