
#1
Nov1312, 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 2^{n+1}1 / 2  1 how do I go about getting this? 2. Relevant equations I tried Ʃ^{n1}_{k=0}ar^{k} = a* 1r^{n} / 1  r with no luck 3. The attempt at a solution = a* 1r^{n+1} / 1  r (which is wrong) Thanks very much! 



#2
Nov1312, 08:59 AM

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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? 



#3
Nov1312, 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 n1 in the summation limit by adding n+1 to the n in the numerator (for the attempted solution)
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! 



#4
Nov1312, 09:43 AM

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Really silly Geometric Progression questionTo 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 ? 



#5
Nov1312, 09:49 AM

P: 174

Cheers 



#6
Nov1312, 10:02 AM

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lateral in thought? 



#7
Nov1312, 10:03 AM

P: 174

Feel free to input anything else...like my question about what if it was 'pi'^i. 



#8
Nov1312, 12:39 PM

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P: 20,990

$$ \sum_{i = 1}^n \pi^{1} = \sum_{i = 1}^n \frac{1}{\pi^i} = \sum_{i = 1}^n \left(\frac{1}{\pi}\right)^i$$ It's still a finite geometric series. 



#9
Nov1312, 01:05 PM

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[tex] a 1  \frac{r^{n+1}}{1}  r,[/tex] but if it was written properly as a(1r^{n+1})/(1r), it would be correct. What makes you think it is wrong? RGV 



#10
Nov1312, 11:57 PM

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[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^{n1}+2^{n}\right)[/itex] 



#11
Nov1412, 07:56 AM

P: 174

Yeah, I'll not neglect the brackets again. 


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