Really silly Geometric Progression question

by toneboy1
Tags: geometric, progression, series
 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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 Quote by toneboy1 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 $\displaystyle \ S=\sum_{i=0}^{n}\left(\frac{1}{2}\right)^{-i}\ .$

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

Then take 2S - S . What do you get?
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 by SammyS Of course that means that $\displaystyle \ S=\sum_{i=0}^{n}\left(2\right)^{i}\ .$
Aaah, YES, that almost seems quite lateral in thought.

 Quote by SammyS 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!

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Really silly Geometric Progression question

 Quote by toneboy1 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: $\displaystyle \ S=1+\sum_{i=1}^{n}\left(2\right)^{i}\ .$

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

Now, what is 2S - S ?
P: 174
 Quote by toneboy1 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 by SammyS I thought that 2S - S might trip you up. To elaborate... Write S as: $\displaystyle \ S=1+\sum_{i=1}^{n}\left(2\right)^{i}\ .$ Write 2S as: $\displaystyle \ 2S=\sum_{i=1}^{n}\left(2\right)^{i}\ \, +\ 2^{n+1}.$ 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
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P: 19,798
 Quote by toneboy1 Aaah, YES, that almost seems quite lateral in thought.
????
lateral in thought?
P: 174
 Quote by Mark44 ????
well I didn't see it.
Feel free to input anything else...like my question about what if it was 'pi'^-i.
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P: 19,798
 Quote by toneboy1 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.
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 Quote by toneboy1 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
$$a 1 - \frac{r^{n+1}}{1} - r,$$
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
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 Quote by toneboy1 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:

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

$\displaystyle =2^1+2^2+2^3+2^4+\dots+2^{n}+2^{n+1}$

$\displaystyle =\sum_{i=1}^{n}\left(2\right)^{i}\ \ +\ \ 2^{n+1}\$
P: 174
 Quote by Mark44 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 by Ray Vickson What you WROTE is wrong because it means $$a 1 - \frac{r^{n+1}}{1} - r,$$ 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 by SammyS Write out a few terms: $\displaystyle \ 2S=2\sum_{i=0}^{n}\left(2\right)^{i}\$$\displaystyle =2\left(2^0+2^1+2^2+2^3+\dots+2^{n-1}+2^{n}\right)$ $\displaystyle =2^1+2^2+2^3+2^4+\dots+2^{n}+2^{n+1}$ $\displaystyle =\sum_{i=1}^{n}\left(2\right)^{i}\ \ +\ \ 2^{n+1}\$
Legend, well explained, thanks!

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