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I've been having a lot of trouble in my calculus class in our section on power series. I understand somewhat but I'm stuck on more complex questions, this is one such question.

My steps to solve this are incorrect but I'm not sure how so. They are the following:

$$\frac{x^5}{27x^3 + 1}$$

Take $x^5$ and separate it from the function

$$x^5 \frac{1}{27x^3 + 1}$$

Rearrange the denominator to match the basic function for the basic power series sum $x^n$

$$x^5 \frac{1}{1 - (-27x^3)}$$

Substitute this into the basic geometric series sum

$$x^5 \sum_{n=0}^\infty (-1)^n (27x^3)^n$$

Which then simplifies to

$$\sum_{n=0}^\infty (-1)^n (27x^{3n+5})$$

There are two possibilities here, that I made a simple mistake in my steps or I don't understand how to do this properly yet.

Any help is appreciated and thank you for reading!

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    $\begingroup$ You forgot the $x$ in your fourth expression , everything else looks fine $\endgroup$ – Shashi Nov 22 '17 at 17:19
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    $\begingroup$ Note that 1/(1-r) is the sum of r^n as n goes from 0 to infinity (if r is between -1 and 1). In the case of this problem, r is -27x^3 and not -27^3. $\endgroup$ – Leonard Blackburn Nov 22 '17 at 17:23
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Hint:

Your mistake is where you

''Substitute this into the basic geometric series sum''

The correct substitution is

$$x^5 \sum_{n=0}^\infty (-1)^n [(3x)^3]^n=x^5 \sum_{n=0}^\infty (-1)^n (3x)^{3n}$$

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  • $\begingroup$ I see, simple algebraic mistake on my part. $\sum_{n=0}^\infty (-1)^n (3x^{3n+5})$ is my answer now, but this is still incorrect apparently. I'll make sure to work on it some more. Mistake likely in my multiplication of $x^5$ $\endgroup$ – Howard P Nov 22 '17 at 17:49
  • $\begingroup$ Works now, the answer I was comparing to wanted me to keep x^5 seperate. Thanks! $\endgroup$ – Howard P Nov 22 '17 at 18:06
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Very clearly worded, I can see that you have a good understanding of the method. You seem to have just dropped the x in the lines following:

Substitute this into the basic geometric series sum

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  • $\begingroup$ Forgetting the x was a mistake I only made in the post, my actual answer includes x and is still incorrect $\endgroup$ – Howard P Nov 22 '17 at 17:28
  • $\begingroup$ Thank you for your compliments though :) $\endgroup$ – Howard P Nov 22 '17 at 17:28
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    $\begingroup$ Have you raised your 27 to the power n? Once you have, this is equivalent to the answer below. $\endgroup$ – bounceback Nov 22 '17 at 17:31

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