sum of the geometric series (2 - 3^n) / 5^n
whats a and r?
its an infinite series by the way
The series is not geometric, as there is no common ratio.
However, it is the difference of two geometric series
(2-3^n)/5^n = 2/5^n - 3^n/5^n = 2(1/5)^n - (3/5)^n
The first series has a = 2/5 and r=1/5
The second series has a = 3/5 and r = 3/5
Now just find the sum of each as usual, and subtract.
so s=0?
no. what is your work?
Just plug in S∞ = a/(1-r)
i did (2/5)^n and i got 5/2. I did (3/5)^n and i got 5/2. 5/2 - 5/2 = 0
(2/5) / (1 - 1/5) = (2/5) / (4/5) = 1/2
(3/5) / (1 - 3/5) = (3/5) / (2/5) = 3/2
1/2 - 3/2 = -1
You showed some work, but you didn't use the values for a and r that I showed you!
how did you get your a and r?
geez - write down the first few terms of each of the geometric sequences.
To find the sum of a geometric series, we need to determine the common ratio (r) and the first term (a).
In the given series (2 - 3^n) / 5^n, let's examine the pattern to find the values of a and r:
The general form of a geometric series is:
a + ar + ar^2 + ar^3 + ...
Comparing the given series with the general form, we can see that:
- The first term is (2 - 3^n), so a = (2 - 3^1) = -1.
- Each term is obtained by multiplying the previous term by a common ratio. In this case, r = (3^n) / 5^n.
So, in the given series, a = -1 and r = (3^n) / 5^n.