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

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The discussion centers on finding the nth term of the series defined by the sum of the first n terms, s, given as 9 - 3^(2-n). Participants clarify that the formula for the sum of a geometric series, s_n = a(1 - r^n)/(1 - r), is applicable only if the series is geometric. Dan corrects the expression to 9 - 3^(2-n) and provides the nth term calculation as a_n = S_n - S_(n-1), resulting in a_n = 18/3^n. The series converges to 9 - 3^(2-n) through summation techniques. The conversation emphasizes the importance of correctly interpreting the series and applying the right formulas.
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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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