How to Simplify a Series with Basic Algebra?

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SUMMARY

The discussion focuses on simplifying the series expression \(\sum (2^{2n}-(-7)^n)/(11^n)\) and its equivalence to \(\sum (4/11)^n - \sum (-7/11)^n\). A key point is the application of the exponent rule \((x^y)^z = x^{yz}\), which clarifies the transformation from \(2^{2n}\) to \(4^n\). Participants emphasize the importance of understanding exponent rules in algebraic simplification, especially in the context of series.

PREREQUISITES
  • Understanding of basic algebraic expressions
  • Familiarity with series and summation notation
  • Knowledge of exponent rules, specifically \((x^y)^z = x^{yz}\)
  • Basic skills in manipulating fractions and common denominators
NEXT STEPS
  • Study the properties of geometric series and their convergence
  • Learn more about algebraic manipulation of series
  • Explore advanced exponent rules and their applications
  • Investigate the implications of series transformations in calculus
USEFUL FOR

Students of mathematics, educators teaching algebra, and anyone interested in mastering series simplification techniques.

kuahji
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[tex]\sum[/tex] (2^(2n)-(-7)^n)/(11^n)

The book has that expression equal to
[tex]\sum[/tex] (4/11)^n - [tex]\sum[/tex] (-7/11)^n

I'm not seeing how the first part changes to (4/11)^n. Wouldn't it be (2^2+2^n) & not 4^n? Or is there something else I'm missing?
 
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Fact:
[tex](x^y)^z = x^{yz}[/tex].
Does that help you out?
 
Crude, nm. 2am & math doesn't mix sometimes. ^_^
 

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