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

[Maatttt0]

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

 

[micromass]

Hi Maatttt0!

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

$$S_n=a\frac{1-r^{n+1}}{1-r}$$

so with an n+1 instead of an n...

 

[Unit]

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 + 3^h + 3^2h + 3^3h + ... + 3^(n-1)h) as S/h = 1 + 3^h + 3^2h + ... + 3^(n-1)h. Noting that this is a geometric progression in 3^h, multiply both sides by 3^h. Then, compute the difference (S/h) - (S/h)3^h. Can you go from there?

 

[Maatttt0]

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$^{h}$ * $\frac{S}{h}$ = 3$^{h}$ + 3$^{2h}$ + ... + 3$^{nh}$

3$^{h}$ * $\frac{S}{h}$ = (3$^{h}$(1-3$^{1}$)) / (1-3$^{h}$)

S = -2h / (1-3$^{h}$) 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.

 

[Unit]

S = -2h / (1-3$^{h}$) 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 :)

 

[Maatttt0]

Yes, that's exactly it! Well done :)

Aha yay! Thank you Unit - muchly appreciated :D