Really silly Geometric Progression question

[toneboy1]

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 Ʃ^n-1_k=0ar^k = 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!

 

[SammyS]

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 Ʃ^n-1_k=0ar^k = 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 \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?

 

[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 (for the attempted solution)

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

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

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]

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 ?

 

[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??

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

 

[Mark44]

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

?
lateral in thought?

 

[toneboy1]

?

well I didn't see it.
Feel free to input anything else...like my question about what if it was 'pi'^-i.

 

[Mark44]

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]

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 Ʃ^n-1_k=0ar^k = 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
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

 

[SammyS]

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}\
​

 

[toneboy1]

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.

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.

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!