Not able to simplify this summation formula?

[musicgold]

Hi,

Please see the attached pdf file. Equation 1 and equation 2 are equivalent.
Can someone please help me understand how to simplify equation 1 to get to equation 2?

Thanks.

 

[Mentallic]

The summation variable is n, so x can be considered a constant hence it can be pulled out the front.

What is

$$1+r+r^2+...+r^n$$

equal to?

 

[musicgold]

The summation variable is n, so x can be considered a constant hence it can be pulled out the front.

What is

$$1+r+r^2+...+r^n$$

equal to?

Please see the new attached file.
This is how far I could go.

 

[HallsofIvy]

The point is that both of your series are geometric series. What you do in your last post is essentially repeating the proof that the sum of the geometric series, $\sum_{n=0}^\infty r^n$ is $\frac{1}{1- r}$ except that you have $r= \frac{1}{\gamma}$.

 

[Mentallic]

The point is that both of your series are geometric series. What you do in your last post is essentially repeating the proof that the sum of the geometric series, $\sum_{n=0}^\infty r^n$ is $\frac{1}{1- r}$ except that you have $r= \frac{1}{\gamma}$.

There is no mention of infinite sums.

musicgold, so what is

$$\sum_{i=1}^{n}\frac{1}{1+y}$$

And hence, let n=10. Also,

$$\frac{1-\frac{1}{\gamma^{n+1}}}{1-\frac{1}{\gamma}}$$

Can be simplified further. At least get rid of the fraction within the denominator.

 

[musicgold]

Please see the attached file.

I think, I am close, but not sure how to get rid of the 'y' in the encircled term.
What am I missing?

 

[Mentallic]

There is no mention of infinite sums.

musicgold, so what is

$$\sum_{i=1}^{n}\frac{1}{1+y}$$

Sorry, this was supposed to be

$$\sum_{i=1}^{n}\frac{1}{(1+y)^i}$$

Please see the attached file.

I think, I am close, but not sure how to get rid of the 'y' in the encircled term.
What am I missing?

We're looking for

$$\sum_{n=1}^{10}\frac{1}{(1+y)^n}$$

while you're finding

$$\sum_{n=0}^{10}\frac{1}{(1+y)^n}$$

In your second attachment when you found

$$s=\sum\frac{1}{\gamma^n}=1+\frac{1}{\gamma}+\frac{1}{\gamma^2}+...$$

You began the sum with n=0 when you should've began with n=1.

 

[musicgold]

Mentallic,

Yes. I was able to solve it with that correction. See attached. Thank you very much.

 

[Mentallic]

Good work