Consider the infinite geometric series ∑∞n=1−4(13)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.

Consider the infinite geometric series n1413n1 In this image the lower limit of the summation notation is n 1 a Write the first four terms of the series b Does class=

Respuesta :

Answer:

See below

Step-by-step explanation:

-4 (1/3)^(n-1)

Part A

For n = 1

-4(1/3)^(1 - 1)

-4(1/3)^0    Anything to the 0 power (except 0) is 1

-4 (1)

-4

n = 2

-4(1/3)^(2 - 1)

-4*(1/3)^1

-4/3

n = 3

- 4(1/3)^2

-4/9

n = 4

-4/(1/3)^3

-4 / 27

Part B

The series converges.

1/3 is between -1 <= 1/3 <= 1

Part C

Formula

Sum = a/(1 - r)

a = - 4

r = 1/3

Sum = -4/(1 - 1/3) = -4//2/3 = - 4/(0.666666666...) = -6

The sum of the converging series with the first term as -4 is -6.

Given to us,

  • Series,  [tex]\sum^{\infty }_{n=1}=-4(\dfrac{1}{3})^{(n-1)}}[/tex]

where, n=1

To find

a.) The first four terms of the series,

[tex]\sum^{\infty }_{n=1}=-4(\dfrac{1}{3})^{(n-1)}}[/tex]

first term,

n=1,

[tex]=-4(\dfrac{1}{3})^{(1-1)}}\\=-4(\dfrac{1}{3})^{(0)}}\\=-4[/tex]

Second term,

n=2,

[tex]=-4(\dfrac{1}{3})^{(2-1)}}\\=-4(\dfrac{1}{3})^{(1)}}\\=-4\times \dfrac{1}{3}\\=\dfrac{-4}{3}[/tex]

Third term,

n=3,

[tex]=-4(\dfrac{1}{3})^{(3-1)}}\\=-4(\dfrac{1}{3})^{(2)}}\\=-4\times \dfrac{1}{9}\\=\dfrac{-4}{9}[/tex]

Fourth term,

n=4,

[tex]=-4(\dfrac{1}{3})^{(4-1)}}\\=-4(\dfrac{1}{3})^{(3)}}\\=-4\times \dfrac{1}{27}\\=\dfrac{-4}{27}[/tex]

To find

b.)  The series diverge or converge,

The series is a converging series as the n value increases the value of the series decreases, because, the more the value of n the smaller number we will get. And, as we can see the n is in the denominator.

Hence, the series is a converging series.

To find

c.)  The sum of the series,

We know that sum of a series is given as [tex]\bold{\dfrac{a}{(1 - r)}}[/tex].

where, a is the first term and r is the common ratio.

In this series,

first term, a = - 4

the common ratio, r = [tex]\dfrac{1}{3}[/tex]

Thus, the sum of the series = [tex]{\dfrac{a}{(1 - r)} = {\dfrac{-4}{(1 - {\frac{1}{3})}} = \dfrac{-4}{\frac{2}{3}}= -6[/tex].

Hence, the sum of the converging series with the first term as -4 is -6.

Learn more about the Series:

https://brainly.com/question/10813422

ACCESS MORE
EDU ACCESS