Find the sum of a geometric progression

In summary, the student is trying to solve for a sum of a geometric progression, but they are unsure of how to reduce the sum to a common denominator. They also question whether or not they have reduced the sum to a common denominator correctly. Additionally, the student has trouble knowing when to cancel out fractions in the numerator and denominator and asks for help with finding the simplest form for the equation.
  • #1
mindauggas
127
0

Homework Statement



(1) [itex]\frac{1}{(1+x^{2})}[/itex]+[itex]\frac{1}{(1+x^{2})^{2}}[/itex]+...+[itex]\frac{1}{(1+x^{2})^{n}}[/itex]

The Attempt at a Solution



(2) [itex]\frac{1}{(1+x^{2})}[/itex]*[itex]\frac{(1+x^{2})^{n-1}}{(1+x^{2})^{n-1}}[/itex]+[itex]\frac{1}{(1+x^{2})^{2}}[/itex]*[itex]\frac{(1+x^{2})^{n-2}}{(1+x^{2})^{n-2}}[/itex]+...+[itex]\frac{1}{(1+x^{2})^{n}}[/itex]*[itex]\frac{(1+x^{2})^{n-n}}{(1+x^{2})^{n-n}}[/itex]

(3) [itex]\frac{(1+x^{2})^{n-1}+(1+x^{2})^{n-2}+...+(1+x^{2})^{n-n}}{(1+x^{2})^{n}}[/itex]

Then I use the formula

[itex]{S(n)=\frac{n(n-1)}{2}}[/itex]

I think i didn't do (2) correctly (reduction to a common denominator). Or did I?
 
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  • #2
What you're doing is unnecessary. You already know that the formula for the sum of a geometric progression is

[tex]S_n=\frac{a(1-r^n)}{1-r}[/tex]

So what is r and a in this case?
 
  • #3
Yeah I probably should have done it this way.

Both a and r are: [itex]\frac{1}{1+x^{2}}[/itex]

Now 3 questions:

(1) Have I reduced sum to a common denominator correctly (previous post eq. (2))?

(2) Should I simplify the eq.:

[tex]S_n=\frac{\frac{1}{1+x^{2}}(1-\frac{1}{1+x^{2}}^n)}{1-\frac{1}{1+x^{2}}}[/tex]

(3) How to know when to leave the exercise, what is the rule of thumb to remember on how much to simplify?
 
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  • #4
mindauggas said:
(2) Should I simplify the eq.:

[tex]S_n=\frac{\frac{1}{1+x^{2}}(1-\frac{1}{1+x^{2}}^n)}{1-\frac{1}{1+x^{2}}}[/tex]

(3) How to know when to leave the exercise, what is the rule of thumb to remember on how much to simplify?

By all means you should simplify. I don't consider a complex fraction to be in simpliest form. Can you simplify so that neither the numerator or denominator contain fractions?
 
  • #5
Yes you have correctly found the common denominator in (2).

You should definitely cancel out the [tex]\frac{1}{1+x^2}[/tex] in both the numerator and denominator. From there, I think it would be fine, if you go further as to get rid of the fraction in the numerator then you're going to have two [tex](1+x^2)^n[/tex] terms and some would argue that it isn't as neat.
 
  • #6
[tex]S_n=\frac{\frac{1}{1+x^{2}}(1-\frac{1}{1+x^{2}}^n)}{1-\frac{1}{1+x^{2}}}[/tex]

I went:

(1) [tex]S_n=\frac{\frac{1}{1+x^{2}}(1-\frac{1}{1+x^{2}}^n)}{\frac{1+x^{2}}{1+x^{2}}-\frac{1}{1+x^{2}}}[/tex]

At the end got:

(2) [tex]S_n=\frac{1-\frac{1}{1+x^{2}}^{n}}{x^{2}}[/tex]

Can I do smth. more (I doesn't seem to me)? Or should i do something different? Is there a way to get rid of the fraction in the numerator?
 
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  • #7
In exactly the same way that you would get rid of the fraction in an expression such as this,

[tex]\frac{1+\frac{1}{x}}{y}[/tex]
 
  • #8
Continuing:

[tex]S_n=\frac{1-\frac{1}{(1+x^{2})^{n}}}{x^{2}}[/tex]

Got:

[tex]S_n=\frac{(1+x^{2})^{n}}{x^{2}(1+x^{2})^{n}-1}[/tex]

Is this correct?
 
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  • #9
Probably should have written a few more steps:

(1) [tex]S_n=\frac{1-\frac{1}{(1+x^{2})^{n}}}{x^{2}}[/tex]

(2) [tex]S_n=\frac{\frac{(1+x^{2})^{n}}{(1+x^{2})^{n}}-\frac{1}{(1+x^{2})^{n}}}{x^{2}}[/tex]

(3) [tex]S_n=\frac{\frac{(1+x^{2})^{n}-1}{(1+x^{2})^{n}}}{x^{2}}[/tex]

(4) [tex]S_n=\frac{x^{2}(1+x^{2})^{n}-1}{(1+x^{2})^{n}}[/tex]

Is this the form that I should leave it in?
 
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  • #10
mindauggas said:
Probably should have written a few more steps:


(4) [tex]S_n=\frac{x^{2}(1+x^{2})^{n}-1}{(1+x^{2})^{n}}[/tex]

Is this the form that I should leave it in?

It should be
[tex]S_n=\frac{(1+x^{2})^{n}-1}{x^{2}(1+x^{2})^{n}}[/tex]
 
  • #11
Yes, my bad.

Thank you all for the help.
 

What is a geometric progression?

A geometric progression is a sequence of numbers where each term is found by multiplying the previous term by a constant number called the "common ratio". The general formula for a geometric progression is an = a1 * r(n-1), where an is the n-th term, a1 is the first term, and r is the common ratio.

How do you find the sum of a geometric progression?

The sum of a geometric progression can be found by using the formula Sn = a1 * (1 - rn) / (1 - r), where Sn is the sum of the first n terms, a1 is the first term, and r is the common ratio. This formula can also be written as Sn = a1 * (1 - rn) / (1 - r).

What is the difference between a finite and infinite geometric progression?

A finite geometric progression has a limited number of terms, while an infinite geometric progression continues indefinitely. In other words, a finite geometric progression has a fixed number of terms, while an infinite geometric progression can have an infinite number of terms.

Can the sum of a geometric progression be negative?

Yes, the sum of a geometric progression can be negative. This happens when the common ratio r is between -1 and 0, and the number of terms n is odd. In this case, the terms of the geometric progression alternate between being positive and negative, resulting in a negative sum.

How can the sum of a geometric progression be used in real life?

The sum of a geometric progression has many real-life applications, such as calculating compound interest in finance, predicting population growth, and analyzing the growth of bacteria or viruses. It can also be used in engineering, physics, and other scientific fields to model and predict exponential growth or decay.

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