Geometric Sequence Sum with Non-Traditional First Term?

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SUMMARY

The discussion clarifies the calculation of the sum of a geometric series when the first term is not equal to 1. Specifically, it establishes that the sum of a geometric series can be expressed as the first term divided by (1 minus the common ratio). For example, the series starting from n=0, \sum_{n=0}^\infty 1/2^n, sums to 2, while the series starting from n=1, \sum_{n=1}^\infty 1/2^n, sums to 1 due to the absence of the initial term. Further examples illustrate that starting from n=2 and n=3 results in sums of 1/2 and 1/4, respectively.

PREREQUISITES
  • Understanding of geometric series and their properties
  • Familiarity with mathematical notation, particularly summation notation
  • Knowledge of the concept of common ratios in sequences
  • Basic algebra skills for manipulating series and equations
NEXT STEPS
  • Study the derivation of the geometric series sum formula
  • Explore variations of geometric series with different first terms
  • Learn about convergence criteria for infinite series
  • Investigate applications of geometric series in real-world scenarios
USEFUL FOR

Students of mathematics, educators teaching series and sequences, and anyone interested in advanced algebraic concepts will benefit from this discussion.

kylera
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In words, the sum of a geometric sequence can be written out to say "the first term divided by (1 minus the common ratio)". Does the first term also apply when the series starts with some other number n other than 1 (like 2 or 3, etc)? In other words, the first term is when n = some other number instead of 1.
 
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Why in the world would you even ask?

\sum_{n=0}^\infty 1/2^n= 1+ 1/2 + 1/4+ \cdot\cdot\cdot
is a geometric series that sums to
\frac{1}{1- 1/2}= 2[/itex]<br /> Why would you think that <br /> \sum_{n=1}^\infty 1/2^n= 1/2 + 1/4+ \cdot\cdot\cdot<br /> sums to the same thing? It is missing the initial 1 so it sums to 2-1= 1.<br /> Similarly<br /> \sum_{n=2}^\infty 1/2^n= 2- 1- 1/2= 1/2<br /> and <br /> \sum_{n= 3}^\infty 1/2^n= 2- 1- 1/2- 1/4= 1/4
 
Well, I'm sorry if the question sounded silly and amateurish, but the book I'm using didn't emphasize that aspect.
 

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