Geometric series

**gopi9** · September 20th 2012, 09:14 AM

Hi everyone,

I am struck in an expansion. Hope someone could help me in this expansion.

The expansion that I have is
1+(2^2)(r^2)+(3^2)(r^4)+(4^2)(r^6)+....... , 1+(2^4)(r^2)+(3^4)(r^4)+(4^4)(r^6)+....... , ......

Can we find geometric progression for this expansion.

Thank you .....

**HallsofIvy** · September 20th 2012, 09:54 AM

Originally Posted by gopi9

Hi everyone,

I am struck in an expansion. Hope someone could help me in this expansion.

The expansion that I have is
1+(2^2)(r^2)+(3^2)(r^4)+(4^2)(r^6)+....... , 1+(2^4)(r^2)+(3^4)(r^4)+(4^4)(r^6)+....... , ......

Can we find geometric progression for this expansion.

Thank you .....

Are these two different problems? If not, what is meaning of the "," between two series? And what do you mean by "the geometric expansion"? Those are NOT geometric serie and you can't just "make" them geometric series.

**gopi9** · September 20th 2012, 09:57 AM

Ya they are 2 different problems.
Can we get a solution for that 1st problem?

**gopi9** · September 20th 2012, 10:00 AM

We have a solution for 1+ar+ar^2+ar^3+.......
that is a/(1-r).
Do we have a solution like this for the problem that i mentioned?

**Soroban** · September 20th 2012, 10:50 AM

Hello, gopi9!

Here is the first one . . .

$S \:=\: 1+2^2r^2+3^2r^4+4^2r^6 \:+\:\hdots$

$\begin{array}{ccccccc}\text{We are given:} &S &=& 1 + 4r^2 + 9r^4 + 16r^6 + 25r^8 + \hdots \\ \text{Multiply by }r^2\!: & r^2S &=& \quad\; r^2 + 4r^4 + 9r^6 + 16 r^8 + \hdots \\ \end{array}$

$\text{Subtract: }\:S - r^2S \;=\;1 + 3r^2 + 5r^4 + 7r^6 + 9 r^8 + \hdots$

$\begin{array}{ccccccc}\text{So we have:} & (1-r^2)S &=& 1 + 3r^2 + 5r^4 + 7r^6 + 9r^8 + \hdots \\ \text{Multiply by }r^2\!: & r^2(1-r^2)S &=& \quad\;r^2 + 3r^4 + 5r^6 + 7r^8 + \hdots \end{array}$

$\text{Subtract: }\:(1-r^2)S - r^2(1-r^2)S \;=\;1 + 2r^2 + 2r^4 + 2r^6 + 2r^8 + \hdots$

$\text{And we have: }\:(1-r^2)^2S \;=\;1 + 2r^2\underbrace{(1 + r^2 + r^4 + r^6 + \hdots)}_{\text{geometric series}}$

$\text{The geometric series has the sum: }\frac{1}{1-r}$

$\text{Hence, we have: }\:(1-r^2)^2S \;=\;1 + 2r^2\left(\frac{1}{1-r}\right) \;=\;\frac{1+r^2}{1-r^2}$

$\text{Therefore: }\:S \;=\;\frac{1+r^2}{(1-r^2)^3}$

**gopi9** · September 20th 2012, 10:55 AM

That looks great..
Thank you so much Soroban..

**gopi9** · September 20th 2012, 01:21 PM

I tried to find the solution for 2nd part in the same way. I tried many iterations but I could not find something that can be taken common(as 2r^2 in the above proof) to make in to geometric series.
The problem is that I have many series like this (2^2, 2^4, 2^6,.....). Can we generalize the solution.
Thanks in advance

**gopi9** · September 20th 2012, 01:30 PM

For the 2nd one the solution that I got is
1+12(r^2)+23(r^4)+24(r^6)(1/(1-(r^2)))= (1+11r^2+11r^4+r^6)/(1-r^2)
but this is so complicated and for higher powers it will be much more complicated..

**johnsomeone** · September 23rd 2012, 03:41 AM

Originally Posted by Soroban

...

$\text{And we have: }\(1-r^2)^2S \;=\;1 + 2r^2\underbrace{(1 + r^2 + r^4 + r^6 + \hdots)}_{\text{geometric series}}$

$\text{The geometric series has the sum: }\frac{1}{1-r}$

$\text{Hence, we have: }\(1-r^2)^2S \;=\;1 + 2r^2\left(\frac{1}{1-r}\right) \;=\;\frac{1+r^2}{1-r^2}$

...

That was a beautiful piece of manipulation. I was admiring it so closely that I found a small transcription error. Small point, but, just in case anyone else is following closely (they should!):

In the 2nd and 3rd lines of the quote, the appearance of $\frac{1}{1-r}$ were obviously intended to be $\frac{1}{1-r^2}$ .

You corrected it by the final equals expresssion on the 3rd line of the quote.

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