A Particular Summation Serie

**mohamedennahdi** · July 24th 2012, 04:50 PM

Hello All,

I am just wondering if this following has a formula:

_{i = 0}∑^{k - 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!

**emakarov** · July 24th 2012, 05:10 PM

This is a geometric progression.

**Plato** · July 24th 2012, 05:23 PM

Originally Posted by mohamedennahdi

I am just wondering if this following has a formula:
_{i = 0}∑^{k - 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 $S = \dfrac{{1 - {{\left( {\frac{x}{y}} \right)}^N}}}{{\left( {1 - \frac{x}{y}} \right)}}$

**mohamedennahdi** · July 24th 2012, 06:30 PM

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!

**mohamedennahdi** · July 24th 2012, 07:25 PM

The Zeno sum

**Prove It** · July 24th 2012, 07:50 PM

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:

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

**emakarov** · July 25th 2012, 02:35 AM

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

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