Find the nth Term of a Series: 9-3^2-n

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Discussion Overview

The discussion revolves around finding the nth term of a series defined by the expression 9-3^2-n. Participants explore the appropriate formulas and methods for deriving the nth term, considering whether the series is geometric and how to interpret the given expression.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested, Mathematical reasoning

Main Points Raised

  • One participant questions the correctness of the expression "9-3^2-n," suggesting it may need parentheses for clarity.
  • Another participant proposes that if the expression is indeed "9-3^{2-n}," then the formula for the sum of a geometric series might apply, specifically mentioning the formula s_n = a(1-r)/(1-r).
  • A later reply clarifies the expression as "9-32-n" and reiterates the previous point about the formula for geometric series.
  • One participant calculates the nth term using the difference of sums, presenting a formula for the nth term as a_n = S_n - S_{n-1} = 9-3^{2-n} - (9-3^{3-n}) = 18/3^n, and checks the summation of the series.

Areas of Agreement / Disagreement

Participants express uncertainty about the initial expression and whether it is correctly typed. There is no consensus on the interpretation of the series or the application of the geometric series formula, indicating multiple competing views remain.

Contextual Notes

There are limitations regarding the clarity of the expression and the assumptions about the series type, which remain unresolved.

ChelseaL
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Given that the sum of the first n terms of series, s, is 9-3^2-n

(i) find the nth term of s.

Do I have to use the formula
sn = a(1-r)/1-r?
 
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"9-3^2-n" is -n? Are you sure it's typed correctly? Use parentheses where appropriate.
 
ChelseaL said:
Do I have to use the formula
sn = a(1-r)/1-r?
If that's supposed to be [math]s_n = a \left ( \frac{1 - r^n}{1-r} \right )[/math] then only if it's a geometric series.

-Dan
 
It's 9-32-n
And what Dan said is what I was trying to say about the formula.
 
$a_n = S_n-S_{n-1} = 9-3^{2-n}-(9-3^{3-n}) = \dfrac{18}{3^n}$

check ...

$\displaystyle \sum_{k=1}^n \dfrac{18}{3^k} = 18 \cdot \sum_{k=1}^n \left(\dfrac{1}{3}\right)^k = \dfrac{6\left[1-\left(\frac{1}{3}\right)^n\right]}{1 - \frac{1}{3}} = 9(1-3^{-n}) = 9 - 3^{2-n}$
 
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