Find the sum of a geometric progression

[mindauggas]

Homework Statement

(1) \frac{1}{(1+x^{2})}+\frac{1}{(1+x^{2})^{2}}+...+\frac{1}{(1+x^{2})^{n}}

The Attempt at a Solution

(2) \frac{1}{(1+x^{2})}*\frac{(1+x^{2})^{n-1}}{(1+x^{2})^{n-1}}+\frac{1}{(1+x^{2})^{2}}*\frac{(1+x^{2})^{n-2}}{(1+x^{2})^{n-2}}+...+\frac{1}{(1+x^{2})^{n}}*\frac{(1+x^{2})^{n-n}}{(1+x^{2})^{n-n}}

(3) \frac{(1+x^{2})^{n-1}+(1+x^{2})^{n-2}+...+(1+x^{2})^{n-n}}{(1+x^{2})^{n}}

Then I use the formula

{S(n)=\frac{n(n-1)}{2}}

I think i didn't do (2) correctly (reduction to a common denominator). Or did I?

 

[Mentallic]

What you're doing is unnecessary. You already know that the formula for the sum of a geometric progression is

S_n=\frac{a(1-r^n)}{1-r}

So what is r and a in this case?

 

[mindauggas]

Yeah I probably should have done it this way.

Both a and r are: \frac{1}{1+x^{2}}

Now 3 questions:

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

(2) Should I simplify the eq.:

S_n=\frac{\frac{1}{1+x^{2}}(1-\frac{1}{1+x^{2}}^n)}{1-\frac{1}{1+x^{2}}}

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

 

[eumyang]

(2) Should I simplify the eq.:

S_n=\frac{\frac{1}{1+x^{2}}(1-\frac{1}{1+x^{2}}^n)}{1-\frac{1}{1+x^{2}}}

(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?

 

[Mentallic]

Yes you have correctly found the common denominator in (2).

You should definitely cancel out the \frac{1}{1+x^2} 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 (1+x^2)^n terms and some would argue that it isn't as neat.

 

[mindauggas]

S_n=\frac{\frac{1}{1+x^{2}}(1-\frac{1}{1+x^{2}}^n)}{1-\frac{1}{1+x^{2}}}

I went:

(1) 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}}}

At the end got:

(2) S_n=\frac{1-\frac{1}{1+x^{2}}^{n}}{x^{2}}

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?

 

[Mentallic]

In exactly the same way that you would get rid of the fraction in an expression such as this,

\frac{1+\frac{1}{x}}{y}

 

[mindauggas]

Continuing:

S_n=\frac{1-\frac{1}{(1+x^{2})^{n}}}{x^{2}}

Got:

S_n=\frac{(1+x^{2})^{n}}{x^{2}(1+x^{2})^{n}-1}

Is this correct?

 

[mindauggas]

Probably should have written a few more steps:

(1) S_n=\frac{1-\frac{1}{(1+x^{2})^{n}}}{x^{2}}

(2) S_n=\frac{\frac{(1+x^{2})^{n}}{(1+x^{2})^{n}}-\frac{1}{(1+x^{2})^{n}}}{x^{2}}

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

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

Is this the form that I should leave it in?

 

[Saitama]

Probably should have written a few more steps:


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

Is this the form that I should leave it in?

It should be
S_n=\frac{(1+x^{2})^{n}-1}{x^{2}(1+x^{2})^{n}}

 

[mindauggas]

Yes, my bad.

Thank you all for the help.