Finding the Sum of an Alternating Geometric Sequence

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Sarah00
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Hi!

If I have a sequence that its first 4 terms are:

30, -31, +32, -32

The pattern is geometric sequence but has alternating signs..

How can I find its sum ..

I know it is composed of 2 sequences ..

However, when I try to separate the 2 sequences .. I get them of different "lengths"

In other words, it is (30+32+34) - (31+33) for 5 terms

but for 4 terms I get:
In other words, it is (30+32) - (31+33)How can I get general formula for both ..

(-1)k helps ! but how! and what about the number of termsThanks!
 
Further to my previous post ..

Handwritten-style typeset derivation of alternating sums of powers of 3 and 9: (3^0 + 3^2) - (3^1 + 3^3) equals -2 times a geometric sum giving -(3^4 - 1)/4, then general formulas for odd n (-(3^n - 1)/4 = (1/4)(-3^n + 1)) and even n (3^n - (3^n - 1)/4 = (1/4)(3^(n+1) + 1)).
 
When ##n## is odd that is ##n=2m+1## you obtain ## \frac{1}{4}(-3^{2m+1}+1)##, when is even ##n=2m## you obtain ## \frac{1}{4}(3^{2m+1}+1)##. In order to obtain both you must have ## \frac{1}{4}((-1)^{n}3^{n+1}+1)## (this is obtained merging the two previous...)
 
Sarah00 said:
Further to my previous post ..

Handwritten-style derivation: (3^0 + 3^2) - (3^1 + 3^3) = -2(3^0 + 3^2) = -2 times ((3^2)^(4/2) - 1)/(3^2 - 1) = -(3^4 - 1)/4; below, for odd n, -(3^n - 1)/4 = (1/4)(-3^n + 1); for even n, 3^n - (3^n - 1)/4 = (1/4)(3^(n+1) + 1).

If you set ##n = 1##, then your formula gives ##S = -1/2##