# Thread: A Particular Summation Serie

1. ## A Particular Summation Serie

Hello All,

I am just wondering if this following has a formula:

i = 0k - 1 Xi/Yi = ?

For example, X = 3, and Y = 2, and k = 5

30/20 + 31/21 + 32/22 + 33/23 + 34/24 = 1 + 1.5 + 2.25 + 3,375 + 5.0625 = 13.1875;

Any hints?

Thanks!

2. ## Re: A Particular Summation Serie

This is a geometric progression.

3. ## Re: A Particular Summation Serie

Originally Posted by mohamedennahdi
I am just wondering if this following has a formula:
i = 0k - 1 Xi/Yi = ?
For example, X = 3, and Y = 2, and k = 5
30/20 + 31/21 + 32/22 + 33/23 + 34/24 = 1 + 1.5 + 2.25 + 3,375 + 5.0625 = 13.1875;
$S = \sum\limits_{k = o}^{N - 1} {{{\left( {\frac{x}{y}} \right)}^k}} = 1 + \left( {\frac{x}{y}} \right) + \cdots + {\left( {\frac{x}{y}} \right)^{N - 1}}$

$\left( {\frac{x}{y}} \right)S = \left( {\frac{x}{y}} \right) + {\left( {\frac{x}{y}} \right)^2} \cdots + {\left( {\frac{x}{y}} \right)^N}$

$\left( {1 - \frac{x}{y}} \right)S = 1 - {\left( {\frac{x}{y}} \right)^N}$

Thus $S = \dfrac{{1 - {{\left( {\frac{x}{y}} \right)}^N}}}{{\left( {1 - \frac{x}{y}} \right)}}$

4. ## Re: A Particular Summation Serie

Wow, this is really a "in a nutshell" response.
However, I can't figure out how we moved from line 2 to line 3.
Can you elaborate if you don't mind?

Thanks!

The Zeno sum

6. ## Re: A Particular Summation Serie

Originally Posted by mohamedennahdi
Wow, this is really a "in a nutshell" response.
However, I can't figure out how we moved from line 2 to line 3.
Can you elaborate if you don't mind?

Thanks!
From Line 2:

\displaystyle \begin{align*} \left(\frac{x}{y}\right)S &= \left(\frac{x}{y}\right) + \left(\frac{x}{y}\right)^2 + \left(\frac{x}{y}\right)^3 + \dots + \left(\frac{x}{y}\right)^{N - 1} + \left(\frac{x}{y}\right)^N \\ 1 + \left(\frac{x}{y}\right)S &= 1 + \left(\frac{x}{y}\right) + \left(\frac{x}{y}\right)^2 + \left(\frac{x}{y}\right)^3 + \dots + \left(\frac{x}{y}\right)^{N - 1} + \left(\frac{x}{y}\right)^N \\ 1 + \left(\frac{x}{y}\right)S &= S + \left(\frac{x}{y}\right)^N \\ 1 - \left(\frac{x}{y}\right)^N &= S - \left(\frac{x}{y}\right)S \\ 1 - \left(\frac{x}{y}\right)^N &= S\left[ 1 - \left(\frac{x}{y}\right) \right] \\ S &= \frac{1 - \left(\frac{x}{y}\right)^N}{1 - \left(\frac{x}{y}\right)} \end{align*}

7. ## Re: A Particular Summation Serie

In other words, line 3 in post #3 is the difference of the first two lines.