Series Question Impossible?

Inertia_Squared

Inertia_Squared

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May 24, 2019
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Hey there, I'm a student from Australia currently completing the year 12 course for my HSC, I'm doing some study before an upcoming exam but I've found a question that has kind of stumped me. We are only expected to know arithmetic and geometric series at this point in the course, but I'm unsure if this question is a geometric series or if there is any way (geometric or not) to solve it.

The question goes as follows: Find the sum of nine terms of the series 3 + 3^(4/3) + 3^(2/3) +...

Now, given the equation for geometric series, the sum should be equal to a(1-r^n)/(1-r), a is fairly simple to find, however, I have no clue how to find r (we're told that r is a constant between all terms, and is found by Tn/Tn-1, however when applying this rule, r does not remain constant - which means that the equation has to be represented in some other way where this rule is able to be true, but I cannot find it at all!) as the only way I can possibly think of representing it is as 3^(sin((n-1)*90)) however that just seems ridiculous, and only fits the first three terms before repeating which isn't what I'd expect.

Any help on the solvability/solution for this question would be greatly appreciated, and thankyou for your time!
 
If r isn't the same between terms then the series is not geometric.

Is there a typo? We could do this if the first term in the series is [math]3^2[/math].

-Dan
 
topsquark said:
If r isn't the same between terms then the series is not geometric.

Is there a typo? We could do this if the first term in the series is [math]3^2[/math].

-Dan
Click to expand...
No typo, I'm assuming there is an error of some sort but I just wasn't sure if it was something I was missing or not. Either way, if you don't think there is a way to find the series (as a geometric) there is probably a typo with the q itself or a printing error (surprisingly not the first time). Thanks for the reassurance, glad to know that it's probably not me, got an exam tomorrow so I was a little worried that I was missing something fundamental ?

Thanks again for your time!
 
Inertia_Squared said:
No typo, I'm assuming there is an error of some sort but I just wasn't sure if it was something I was missing or not. Either way, if you don't think there is a way to find the series (as a geometric) there is probably a typo with the q itself or a printing error (surprisingly not the first time). Thanks for the reassurance, glad to know that it's probably not me, got an exam tomorrow so I was a little worried that I was missing something fundamental ?

Thanks again for your time!
Click to expand...
If I were to do this problem, assuming that there is a "pattern" and the sum of "first consecutive nine terms" are in question, I would do it by brute-force (since small finite terms)

3^(3/3) + 3^(4/3) + 3^(2/3) + 3^(1/3) + 3^(0/3) + 3^(-1/3) + 3^(-2/3) + 3^(-3/3) + 3^(-4/3)

=[3^(3/3) + 3^(-3/3)] + [3^(4/3) + 3^(-4/3)] + [3^(2/3) + 3^(-2/3)] + [3^(1/3) + 3^(-1/3)] + 3^(0/3) .... continue...
 
If @Subhotosh Khan is correct about how the series progresses, then another method would be to re-order the terms. For example...

3^(3/3) + 3^(4/3) + 3^(2/3) + 3^(1/3) + 3^(0/3) + 3^(-1/3) + 3^(-2/3) + 3^(-3/3) + 3^(-4/3)

move the red term to the left, and we have a recognisable geometric progression...

3^(4/3) + 3^(3/3) + 3^(2/3) + 3^(1/3) + 3^(0/3) + 3^(-1/3) + 3^(-2/3) + 3^(-3/3) + 3^(-4/3))
 
haha thanks for the input, its great to see the lore of the forums unfold as I'm getting my questions answered :p
 
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