Sum of Geometric Series: What Am I Doing Wrong?

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Discussion Overview

The discussion revolves around the calculation of the sum of a geometric series, specifically the series \(\sum_{n=1}^\infty 9(2/3)^n\). Participants explore the correct values for the first term and the common ratio, as well as the implications of the starting index on the sum.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant initially states that the sum can be expressed as \(a/(1-r)\) with \(a=9\) and \(r=2/3\), but expresses confusion over the correct value of \(a\), suggesting it should be 6 instead.
  • Another participant confirms that taking \(a(1)\) as 6 is correct.
  • A different participant points out that the beginning index of the sum is crucial for determining the correct formula to use, providing the relationship between sums starting from different indices.
  • Participants discuss the derivation of the sum from \(n=1\) to \(\infty\) and provide calculations that lead to a result of 18, using both the factorization of \(r\) and the subtraction of the first term from the sum starting at \(n=0\).
  • There is a request for clarification on how to factor out \(r\) in the context of the geometric series.
  • Another participant explains that factoring \(r\) involves applying the distributive law to the series terms.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct value of \(a\) initially, but there is agreement on the importance of the starting index for the sum. The calculations provided lead to a common result of 18, though the method of arriving at that result is debated.

Contextual Notes

The discussion highlights the dependency on the starting index of the series and the definitions of the first term and common ratio, which are not fully resolved. There are also unresolved mathematical steps in the derivation of the sum.

esvion
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I have to find the sum of [tex]\sum9(2/3)^n[/tex] and I get a/1-r where a=9 and r=2/3...but I know a=6 and not 9. Can someone point out to me what I am doing wrong? The sum is from n=1 to infinity.

Thanks!

EDIT: I am thinking I take a(1) which is 6 as the a in a/(1-r), is this correct?
 
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esvion said:
I am thinking I take a(1) which is 6 as the a in a/(1-r), is this correct?

Yep!
 
esvion said:
I have to find the sum of [tex]\sum9(2/3)^n[/tex] and I get a/1-r where a=9 and r=2/3...but I know a=6 and not 9. Can someone point out to me what I am doing wrong? The sum is from n=1 to infinity.

Thanks!

EDIT: I am thinking I take a(1) which is 6 as the a in a/(1-r), is this correct?
You left out important information! What is the beginning index for your sum?

[tex]\sum_{n=0}^\infty a r^n= \frac{a}{1- r}[/itex]<br /> <br /> but <br /> [tex]\sum_{n=1}^\infty a r^n= r \sum_{n=0}^\infty a r^n[/tex]<br /> where I have factored out an "r" to reduce the first index to 0,<br /> and<br /> [tex]\sum_{n=1}^\infty a r^n= \left(\sum_{n=0}^\infty a r^n\right) - a[/tex]<br /> where I have subracted off the [itex]ar^0[/itex] term.<br /> <br /> In particular, with a= 9, r= 2/3, <br /> [tex]\sum_{n=1}^\infty 9 (2/3)^n= (2/3)\sum_{n=0}^\infty 9 (2/3)^n[/tex]<br /> [tex]= \frac{2}{3}\frac{9}{1- 2/3}= \frac{6}{1- 2/3}= 18[/tex]<br /> <br /> or <br /> [tex]\sum_{n=1}^\infty 9(2/3)^n= \left(\sum_{n=0}^\infty 9 (2/3)^n\right)- 9[/tex]<br /> [tex]= \frac{9}{1- 2/3}- 9= 27- 9= 18[/tex][/tex]
 
[tex] \sum_{n=1}^\infty a r^n= r \sum_{n=0}^\infty a r^n[/tex]

hmm, may i know how to factorise that r please?
 
annoymage said:
[tex] \sum_{n=1}^\infty a r^n= r \sum_{n=0}^\infty a r^n[/tex]

hmm, may i know how to factorise that r please?
It's just the "distributive law": [itex]ar+ ar^2+ ar^3+ \cdot\cdot\cdot= r(a+ ar+ ar^2+ \cdot\cdot\cdot[/itex]
 
owh, thank you very much
 

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