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

[ChelseaL]

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?

 

[Greg]

"9-3^2-n" is -n? Are you sure it's typed correctly? Use parentheses where appropriate.

 

[topsquark]

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

 

[ChelseaL]

It's 9-3^2-n
And what Dan said is what I was trying to say about the formula.

 

[skeeter]

$a_n = S_n-S_{n-1} = 9-3^{2-n}-(9-3^{3-n}) = \dfrac{18}{3^n}$

check ...

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