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