Solving Geometric Progression: Sum of h(1 + 3^h + 3^2h + ... + 3^(n-1)h)

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In summary, the conversation was about solving a geometric progression problem involving finding the sum of a series. The formula Sn = a(1-r^n)/(1-r) was used, but it was later suggested to use the technique of computing the difference between S and rS and re-arranging the equation. This led to the correct solution of S = -2h / (1-3^{h}).
  • #1
Maatttt0
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Homework Statement



This isn't the whole question, I understand the prior parts but somehow stuck on the "easy" part :( Need to solve a geometric progession problem.. find the sum of:

h(1 + 3^h + 3^2h + ... + 3^(n-1)h)

Where nh = 1

The sum should equal to (2h)/((3^h) -1) which is given as is a prove question

Homework Equations



N/A

The Attempt at a Solution



h(1 + 3^h + 3^2h + ... + 3^(n-1)h) --- this is the list

(h(1-(3^h)^(n-1)))/(1-3^h) --- using the formula Sn = a(1-r^n)/(1-r)

simplify slightly (using nh = 1)

= (h(1-3^(1-h)))/(1-3^h)

times top and bottom by -1 to match the denominator of answer

= (h(3^(1-h) - 1))/((3^h) -1)

and now I can't seem to get the final answer out.

Hopefully I've included enough info for you guys to understand what I'm going on about lol - and apologise for the excess of brackets :S

Thank you in advance :D
 
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  • #2
Hi Maatttt0! :smile:

Maatttt0 said:

Homework Statement



This isn't the whole question, I understand the prior parts but somehow stuck on the "easy" part :( Need to solve a geometric progession problem.. find the sum of:

h(1 + 3^h + 3^2h + ... + 3^(n-1)h)

Where nh = 1

The sum should equal to (2h)/((3^h) -1) which is given as is a prove question

Homework Equations



N/A

The Attempt at a Solution



h(1 + 3^h + 3^2h + ... + 3^(n-1)h) --- this is the list

(h(1-(3^h)^(n-1)))/(1-3^h) --- using the formula Sn = a(1-r^n)/(1-r)

That is not the correct formula, is it? Isn't the correct formula

[tex]S_n=a\frac{1-r^{n+1}}{1-r}[/tex]

so with an n+1 instead of an n...
 
  • #3
Advice: don't memorize any geometric series formula. Instead, try this:

If you have a geometric progression in r, i.e. S = a + ar + ar2 + ar3 + ... + arn, compute S - rS = S(1 - r) and then re-arrange. This leads you exactly to the formula micromass posted, but it really helps you see what's going on. I prefer to memorize a technique than a formula.

For your problem, you should start by re-writing S = h(1 + 3h + 32h + 33h + ... + 3(n-1)h) as S/h = 1 + 3h + 32h + ... + 3(n-1)h. Noting that this is a geometric progression in 3h, multiply both sides by 3h. Then, compute the difference (S/h) - (S/h)3h. Can you go from there?
 
  • #4
Hey guys,

Thank you for both of the replies :)

micromass; I'm didn't think it was n+1 :S my formula booklet just has n by itself.

Unit; I got the answer to come out I think xD

3[itex]^{h}[/itex] * [itex]\frac{S}{h}[/itex] = 3[itex]^{h}[/itex] + 3[itex]^{2h}[/itex] + ... + 3[itex]^{nh}[/itex]

3[itex]^{h}[/itex] * [itex]\frac{S}{h}[/itex] = (3[itex]^{h}[/itex](1-3[itex]^{1}[/itex])) / (1-3[itex]^{h}[/itex])

S = -2h / (1-3[itex]^{h}[/itex]) then just times top and bottom by -1.

Is this what you were suggesting? Hinting to divide through by h and times through by 3^h was useful.
 
  • #5
Maatttt0 said:
S = -2h / (1-3[itex]^{h}[/itex]) then just times top and bottom by -1.

Is this what you were suggesting? Hinting to divide through by h and times through by 3^h was useful.

Yes, that's exactly it! Well done :)
 
  • #6
Unit said:
Yes, that's exactly it! Well done :)

Aha yay! Thank you Unit - muchly appreciated :D
 

1. What is a geometric progression?

A geometric progression is a sequence of numbers where each term is obtained by multiplying the previous term by a fixed number. For example, the sequence 2, 6, 18, 54 is a geometric progression with a common ratio of 3.

2. 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 given by:
an = a1 * rn-1
where an is the nth term, a1 is the first term, and r is the common ratio.

3. How can I determine if a sequence is a geometric progression?

A sequence is a geometric progression if the ratio between any two consecutive terms is constant. This means that if you divide any term by the previous term, you will get the same number each time. You can also check by using the formula for finding the nth term and seeing if it holds true for all terms in the sequence.

4. What is the sum of a geometric progression?

The sum of a finite geometric progression can be calculated using the formula:
Sn = a * (1 - rn) / (1 - r)
where Sn is the sum of the first n terms, a is the first term, and r is the common ratio.

5. How is geometric progression used in real life?

Geometric progression is used in various fields such as finance, physics, and biology. In finance, it is used to calculate compound interest and in physics, it is used to model exponential growth or decay. In biology, it can be used to model the growth of a population. It is also used in computer algorithms for efficient data storage and retrieval.

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