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

Check ...In summary, the conversation is discussing finding the nth term of a series, denoted as s, given that the sum of the first n terms is represented by 9-3^2-n. There is confusion about the notation used, with one person suggesting the use of the formula sn = a(1-r)/1-r and another person questioning if it is a geometric series. Ultimately, it is determined that the correct formula to use is $a_n = S_n-S_{n-1} = 9-3^{2-n}-(9-3^{3-n}) = \dfrac{18}{3^n}$, and after simplifying, it is equal to 9 - 3^2-n
  • #1
ChelseaL
22
0
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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  • #2
"9-3^2-n" is -n? Are you sure it's typed correctly? Use parentheses where appropriate.
 
  • #3
ChelseaL said:
Do I have to use the formula
sn = a(1-r)/1-r?
If that's supposed to be \(\displaystyle s_n = a \left ( \frac{1 - r^n}{1-r} \right )\) then only if it's a geometric series.

-Dan
 
  • #4
It's 9-32-n
And what Dan said is what I was trying to say about the formula.
 
  • #5
$a_n = S_n-S_{n-1} = 9-3^{2-n}-(9-3^{3-n}) = \dfrac{18}{3^n}$

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

What is the general formula for finding the nth term of a series?

The general formula for finding the nth term of a series is an = a1 + (n-1)d, where an is the nth term, a1 is the first term, and d is the common difference.

How do I determine the first term and common difference of a series?

The first term can be determined by looking at the first number in the series. The common difference can be found by subtracting the first term from the second term, or any subsequent term from the previous term.

How do I apply the formula to the series 9, 3, 2, -n?

To apply the formula to this series, you will need to determine the first term and common difference. In this case, the first term is 9 and the common difference is -6 (9 - 3 = 6, 3 - 2 = 1, 2 - (-n) = 2 + n = 1). Then, plug these values into the formula an = 9 + (n-1)(-6) and simplify to get the nth term.

How can I check if my answer for the nth term is correct?

You can check your answer by plugging in the value of n into the formula and seeing if it matches the corresponding term in the series. For example, if you found that the 5th term is 27, you can plug in n=5 into the formula and see if you get 27 as the answer.

Can the formula be used for any type of series?

Yes, the formula can be used for arithmetic and geometric series, as long as the series follows a pattern and has a constant common difference or ratio.

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