Really silly Geometric Progression question

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In summary, the conversation discussed the solution to a summation problem where the answer is 2n+1-1 / 2 - 1. The equation used to solve the problem was a* 1-rn+1 / 1 - r, with the correction that it should be written properly as a(1-rn+1)/(1-r). The conversation also touched on the topic of manipulating equations and the concept of a finite geometric series.
  • #1
toneboy1
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Homework Statement


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?

Homework Equations



I tried Ʃn-1k=0ark = a* 1-rn / 1 - r

with no luck

The Attempt at a Solution



= a* 1-rn+1 / 1 - r (which is wrong)Thanks very much!
 
Last edited:
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  • #2
toneboy1 said:

Homework Statement


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?

Homework Equations



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

with no luck

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?
 
  • #3
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)

SammyS said:
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.
SammyS said:
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!
 
Last edited:
  • #4
toneboy1 said:
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 ?
 
  • #5
toneboy1 said:
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??
SammyS said:
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
 
  • #6
toneboy1 said:
Aaah, YES, that almost seems quite lateral in thought.
?
lateral in thought?
 
  • #7
Mark44 said:
?

well I didn't see it.
Feel free to input anything else...like my question about what if it was 'pi'^-i.
 
  • #8
toneboy1 said:
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.
 
  • #9
toneboy1 said:

Homework Statement


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?

Homework Equations



I tried Ʃn-1k=0ark = a* 1-rn / 1 - r

with no luck

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
 
  • #10
toneboy1 said:
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]
 
  • #11
Mark44 said:
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.

Ray Vickson said:
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.

SammyS said:
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!
 

1. What is a geometric progression?

A geometric progression is a sequence of numbers where each term is found by multiplying the previous term by a constant ratio. For example, in the sequence 2, 6, 18, 54, the ratio between each term is 3.

2. How is a geometric progression different from an arithmetic progression?

In an arithmetic progression, the difference between each term is a constant value. In a geometric progression, the ratio between each term is a constant value.

3. What is the formula for finding the nth term in a geometric progression?

The formula for finding the nth term in a geometric progression is an = a1 * rn-1, where an is the nth term, a1 is the first term, and r is the common ratio.

4. How can geometric progressions be applied in real life?

Geometric progressions can be found in many real-life scenarios, such as population growth, interest rates, and compound interest. For example, when a population grows at a constant percentage each year, it can be modeled using a geometric progression.

5. What are some common mistakes to avoid when working with geometric progressions?

Some common mistakes to avoid when working with geometric progressions include forgetting to account for the common ratio, using the wrong formula to find the nth term, and not checking for the convergence or divergence of the progression. It is important to carefully follow the steps and double-check your work when working with geometric progressions.

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