# Homework Help: Find the sum of a geometric progression

1. Aug 17, 2011

### mindauggas

1. The problem statement, all variables and given/known data

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

3. 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?

Last edited: Aug 17, 2011
2. Aug 17, 2011

### 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?

3. Aug 17, 2011

### 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?

Last edited: Aug 17, 2011
4. Aug 17, 2011

### eumyang

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. Aug 17, 2011

### 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.

6. Aug 17, 2011

### 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?

Last edited: Aug 17, 2011
7. Aug 17, 2011

### Mentallic

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

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

8. Aug 17, 2011

### 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?

Last edited: Aug 17, 2011
9. Aug 18, 2011

### 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?

Last edited: Aug 18, 2011
10. Aug 18, 2011

### Saitama

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

11. Aug 18, 2011