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Really silly Geometric Progression question

by toneboy1
Tags: geometric, progression, series
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toneboy1
#1
Nov13-12, 08:04 AM
P: 174
1. The problem statement, all variables and given/known data
I should know this (it's been a few years) but can't seem to get the answer for: Ʃ (from i = 0 to n) 0.5-i the answer is apparently 2n+1-1 / 2 - 1

how do I go about getting this?

2. Relevant equations

I tried Ʃn-1k=0ark = a* 1-rn / 1 - r

with no luck

3. The attempt at a solution

= a* 1-rn+1 / 1 - r (which is wrong)


Thanks very much!
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SammyS
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Nov13-12, 08:59 AM
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Quote Quote by toneboy1 View Post
1. The problem statement, all variables and given/known data
I should know this (it's been a few years) but can't seem to get the answer for: Ʃ (from i = 0 to n) 0.5-i the answer is apparently (2n+1-1) /( 2 - 1)

how do I go about getting this?

2. Relevant equations

I tried Ʃn-1k=0ark = a* (1-rn) / (1 - rn)

with no luck

3. The attempt at a solution

= a*(1-rn) / (1 - rn) (which is wrong)

Thanks very much!
First of all, you need to use parentheses to say what (I hope) you intend to say.

Start by letting [itex]\displaystyle \ S=\sum_{i=0}^{n}\left(\frac{1}{2}\right)^{-i}\ .[/itex]

Of course that means that [itex]\displaystyle \ S=\sum_{i=0}^{n}\left(2\right)^{i}\ .[/itex]

Then take 2S - S . What do you get?
toneboy1
#3
Nov13-12, 09:07 AM
P: 174
Thanks for the reply SammyS, sorry about the lack of brackets, you were almost correct except I didn't have an n in the exponent of the denominator and I made up for the n-1 in the summation limit by adding n+1 to the n in the numerator (for the attempted solution)

Quote Quote by SammyS View Post
Of course that means that [itex]\displaystyle \ S=\sum_{i=0}^{n}\left(2\right)^{i}\ .[/itex]
Aaah, YES, that almost seems quite lateral in thought.


Quote Quote by SammyS View Post
Then take 2S - S . What do you get?
H'mm, not sure I'm following...'S'...?
Anyway, that equation I tried before now works if I'm not mistaken.

BUT this begs the question, what if rather than being a half (0.5) and being able to remove the numerator making the exponent positive, what if it was like pi ? what would we do then??

Thanks!

SammyS
#4
Nov13-12, 09:43 AM
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Really silly Geometric Progression question

Quote Quote by toneboy1 View Post
Thanks for the reply SammyS, sorry about the lack of brackets, you were almost correct except I didn't have an n in the exponent of the denominator and I made up for the n-1 in the summation limit by adding n+1 to the n in the numerator.

Aaah, YES, that almost seems quite lateral in thought.

H'mm, not sure I'm following...'S'...?
Anyway, that equation I tried before now works if I'm not mistaken.
I thought that 2S - S might trip you up.

To elaborate...

Write S as: [itex]\displaystyle \ S=1+\sum_{i=1}^{n}\left(2\right)^{i}\ .[/itex]

Write 2S as: [itex]\displaystyle \ 2S=\sum_{i=1}^{n}\left(2\right)^{i}\ \, +\ 2^{n+1}.[/itex]

Now, what is 2S - S ?
toneboy1
#5
Nov13-12, 09:49 AM
P: 174
Quote Quote by toneboy1 View Post
BUT this begs the question, what if rather than being a half (0.5) and being able to remove the numerator making the exponent positive, what if it was like pi ? what would we do then??
Quote Quote by SammyS View Post
I thought that 2S - S might trip you up.

To elaborate...

Write S as: [itex]\displaystyle \ S=1+\sum_{i=1}^{n}\left(2\right)^{i}\ .[/itex]

Write 2S as: [itex]\displaystyle \ 2S=\sum_{i=1}^{n}\left(2\right)^{i}\ \, +\ 2^{n+1}.[/itex]

Now, what is 2S - S ?
would it just be 2S= 2n+1 - 1? Although I don't see how 2S was equal to the expression you said in the first place...?

Cheers
Mark44
#6
Nov13-12, 10:02 AM
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Quote Quote by toneboy1 View Post
Aaah, YES, that almost seems quite lateral in thought.
????
lateral in thought?
toneboy1
#7
Nov13-12, 10:03 AM
P: 174
Quote Quote by Mark44 View Post
????
well I didn't see it.
Feel free to input anything else...like my question about what if it was 'pi'^-i.
Mark44
#8
Nov13-12, 12:39 PM
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Quote Quote by toneboy1 View Post
well I didn't see it.
Feel free to input anything else...like my question about what if it was 'pi'^-i.
I don't see how that would make any difference.

$$ \sum_{i = 1}^n \pi^{-1} = \sum_{i = 1}^n \frac{1}{\pi^i} = \sum_{i = 1}^n \left(\frac{1}{\pi}\right)^i$$

It's still a finite geometric series.
Ray Vickson
#9
Nov13-12, 01:05 PM
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Quote Quote by toneboy1 View Post
1. The problem statement, all variables and given/known data
I should know this (it's been a few years) but can't seem to get the answer for: Ʃ (from i = 0 to n) 0.5-i the answer is apparently 2n+1-1 / 2 - 1

how do I go about getting this?

2. Relevant equations

I tried Ʃn-1k=0ark = a* 1-rn / 1 - r

with no luck

3. The attempt at a solution

= a* 1-rn+1 / 1 - r (which is wrong)


Thanks very much!
What you WROTE is wrong because it means
[tex] a 1 - \frac{r^{n+1}}{1} - r,[/tex]
but if it was written properly as a(1-rn+1)/(1-r), it would be correct. What makes you think it is wrong?

RGV
SammyS
#10
Nov13-12, 11:57 PM
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Quote Quote by toneboy1 View Post
would it just be 2S= 2n+1 - 1? Although I don't see how 2S was equal to the expression you said in the first place...?

Cheers
Write out a few terms:

[itex]\displaystyle \ 2S=2\sum_{i=0}^{n}\left(2\right)^{i}\ [/itex]
[itex]\displaystyle =2\left(2^0+2^1+2^2+2^3+\dots+2^{n-1}+2^{n}\right)[/itex]

[itex]\displaystyle =2^1+2^2+2^3+2^4+\dots+2^{n}+2^{n+1}[/itex]

[itex]\displaystyle =\sum_{i=1}^{n}\left(2\right)^{i}\ \ +\ \ 2^{n+1}\ [/itex]
toneboy1
#11
Nov14-12, 07:56 AM
P: 174
Quote Quote by Mark44 View Post
I don't see how that would make any difference.

$$ \sum_{i = 1}^n \pi^{-1} = \sum_{i = 1}^n \frac{1}{\pi^i} = \sum_{i = 1}^n \left(\frac{1}{\pi}\right)^i$$

It's still a finite geometric series.
How right you are, I was over complicating it in my head, thanks.

Quote Quote by Ray Vickson View Post
What you WROTE is wrong because it means
[tex] a 1 - \frac{r^{n+1}}{1} - r,[/tex]
but if it was written properly as a(1-rn+1)/(1-r), it would be correct. What makes you think it is wrong?

RGV
I was unnecessarily trying to minipulate it and ended up putting in the wrong value, like a circle through a square hole. Rather than just flipping it all.
Yeah, I'll not neglect the brackets again.

Quote Quote by SammyS View Post
Write out a few terms:

[itex]\displaystyle \ 2S=2\sum_{i=0}^{n}\left(2\right)^{i}\ [/itex]
[itex]\displaystyle =2\left(2^0+2^1+2^2+2^3+\dots+2^{n-1}+2^{n}\right)[/itex]

[itex]\displaystyle =2^1+2^2+2^3+2^4+\dots+2^{n}+2^{n+1}[/itex]

[itex]\displaystyle =\sum_{i=1}^{n}\left(2\right)^{i}\ \ +\ \ 2^{n+1}\ [/itex]
Legend, well explained, thanks!


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