(a) really? Just plug in the values for n
-4/3^0, -4/3^1, -4/3^2, -4/3^3
(b) this is just a GP with r = 1/3
(c) S = a/(1-r) = -4/(1 - 1/3) = -6
Consider the infinite geometric series
∑^(∞)_(n=1) −4(1/3)^n−1
. In this image, the lower limit of the summation notation is "n = 1".
a. Write the first four terms of the series.
b. Does the series diverge or converge?
c. If the series has a sum, find the sum.
2 answers
The series is
∑^(∞)_(n=1) −4(1/3)^n−1 = -4 -4/3 -4/3^2 - 4/3^3 - 4/3^4
(first four terms are listed)
b. The series converges because this is a geometric series.
4= 1/3 <1
the sum is -6
You can find the explanation for these on brainly
Since you can't post URLs here just search
Consider the infinite geometric series ∑∞ n=1 -4(1/3)^n-1
a. Write the first four terms of the series.
b. Does the series diverge or converge.
c. If the series has a sum, find the sum.
∑^(∞)_(n=1) −4(1/3)^n−1 = -4 -4/3 -4/3^2 - 4/3^3 - 4/3^4
(first four terms are listed)
b. The series converges because this is a geometric series.
4= 1/3 <1
the sum is -6
You can find the explanation for these on brainly
Since you can't post URLs here just search
Consider the infinite geometric series ∑∞ n=1 -4(1/3)^n-1
a. Write the first four terms of the series.
b. Does the series diverge or converge.
c. If the series has a sum, find the sum.
2 answers