MHB How Do You Solve a Geometric Sum with Alternating Signs?

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SUMMARY

The discussion focuses on solving a geometric sum with alternating signs, specifically the series S = 3 - 3/2 + 3/4 - 3/8 + 3/16 - 3/32 + ... - 3/128. The common ratio identified is r = -1/2, with the first term a = 3. The sum can be calculated using the geometric sum formula, S = a * (1 - r^(n+1)) / (1 - r), where n is the number of terms, leading to S = 3 * (1 - (-1/2)^(8)) / (1 - (-1/2)).

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  • Understanding of geometric series and their properties
  • Familiarity with the geometric sum formula
  • Basic knowledge of rational numbers and integer manipulation
  • Ability to work with alternating series
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  • Study the derivation and application of the geometric sum formula
  • Learn about the convergence of alternating series
  • Explore examples of geometric series with different common ratios
  • Practice solving problems involving rational numbers and series
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Students in mathematics, educators teaching series and sequences, and anyone interested in mastering geometric sums and their applications.

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Hey!

I'm stuck again and not sure how to solve this question been at it for a few hours. Any help is appreciated as always.

Q: (1) Let the sum S = 3- 3/2 + 3/4 - 3/8 + 3/16 - 3/32 +...- 3/128. Determine integers a , n and a rational number k so that...(Image)

r/askmath - Summation and geometric sums

(2 )And then calculate S using the geometric sum formula.

Thank you!
 
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common ratio is $r = -\dfrac{1}{2}$

note $128 = 2^7$

first term is $a = 3$

$\displaystyle S = 3 \sum_{j=0}^7 \left(-\dfrac{1}{2}\right)^j$

you can calculate the sum ...
 
Given that it is a "geometric sum", a+ ar+ ar^2+ ...,, you can determine r, the "common ratio" by just dividing the second term by the first: ar/a= r. In this problem that is (-3/2)/3= -1/2.
 

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