Calculating Sum of Series: 1, (4-a)x, (7-a^2)x^2...

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SUMMARY

The discussion focuses on calculating the sum of the series defined by the terms 1, (4-a)x, (7-a^2)x^2, (10-a^3)x^3, and (13-a^4)x^4. The correct approach involves expanding the series and utilizing power series techniques. The sum to n terms, denoted as Sn, can be expressed as a combination of power series and geometric series, specifically Sn = (power series of 1, 4x, 7x^2...) - ax(1 - ax^(n-1))/(1 - ax). The logical breakdown of the series is essential for accurate calculation.

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Procrastinate
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I just had this question on my exam and I was wondering whether my method of calculating it was right:

I was meant to find sum to n terms of this:

1, (4-a)x, (7-a^2)x^2, (10-a^3)x^3, (13-a^4)x^4...

Would the correct way to go about it be expand it:

1, 4x-ax. 7x^2-a^2x^2...

and then it would be:

Sn = (power series of 1, 4x, 7x^2...) - ax(1-ax^(n-1)/(1-ax)
 
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Hi Procrastinate! :smile:

(try using the X2 tag just above the Reply box :wink:)

The first thing to do with any question like this (in my opinion, anyway) is to write it logically …

it's ∑ (3n + 1 - an)xn

= ∑ xn + 3∑ nxn - ∑(ax)n:wink:
 

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