Sum of Getometric Sequence with alternating signs

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The discussion focuses on finding the sum of a geometric sequence with alternating signs, specifically 5^2 - 5^3 + 5^4 - ... + (-1)^k*5^k for k ≥ 2. The initial approach involves recognizing the sum of positive terms and attempting to adjust for the alternating signs. Two methods are proposed: one involves factoring out (-5)^2 and applying the geometric sum formula, while the other suggests using the formula for the entire sequence and subtracting the first two terms. The calculations reveal that the sum can be derived correctly by either method, confirming the importance of handling the starting index appropriately. Understanding the manipulation of geometric series is key to solving such problems.
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Homework Statement



5^2 - 5^3 + 5^4 - ... + (-1)^k*5^k whre k is an integer with k >= 2

Homework Equations




The Attempt at a Solution



I know (5^(k-1) - 5^2)/2 gives you the sum if they were all positive. I tried multiplying it by (-1)^k or something but that just changes the sign. I wish I could give you more but I can't.
 
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A geometric sequence is \sum_{n=0}^N ar^n. And the sum is:
\frac{1- r^{N+1}}{1- r}
r does not have to be positive. Your sequence has a= 1, r= -5.
 
\frac{1- (-5)^{2+1}}{1- (-5)} = 21 though and not 25
 
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The sum that HallsofIvy gave includes (-5)0 and (-5)1
 
There are two ways to handle the fact that your sum starts with r^2 rather than r^0= 1.

1) Factor out an r^2 (-5)^2+ (-5)^3\cdot\cdot\cdot+ (-5)^k= (-5)^2(1+ (-5)+ \cdot\cdot\cdot+ (-5)^{k-2})

Use the formula I gave with n= k- 2 and then multiply by (-5)^2= 25.

2) Use the formula with n= k and then subtract of (-5)^0+ (-5)^1= 1- 6= -4.
 
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