NaLuShipper4Life
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consider the infinite geometric series infinity sigma n=1 -4(1/3)^n-1. In this, the lower limit of the summation notion 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.

Respuesta :

Answer:

a) The first four terms are : -3 , -1/3 , 5/9 , 23/27

b) The series converges

c) S∞ = (1)(n - 2) = ∞

Step-by-step explanation:

a) ∵ αn = 1 - 4(1/3)^n-1

∴ If n = 1 ⇒ α1 = 1 - 4(1/3)^0 = -3

∴ If n = 2 ⇒ α2 = 1 - 4(1/3)^1 = -1/3

∴ If n = 3 ⇒ α3 = 1 - 4(1/3)^2 = 5/9

∴ If n = 4 ⇒ α4 = 1 - 4(1/3)^3 = 23/27

b) The series converges because with large number of n

it approached to 1

Note: If IrI < 1 then ⇒ converge

c) S∞ = (1)(n - 2) ⇒ If n go to ∞

S∞ = ∞

MrRoyal

The true statements are:

  • The first four terms are -4, -4/3, -4/9 and -4/27
  • The series converges
  • The sum of the series is -6

The summation notation is given as:

[tex]\sum\limits^{\infty}_{n =1} -4(\frac 13)^{n-1[/tex]

When n = 1, we have:

[tex]T_1 = -4(\frac 13)^{1-1} = -4[/tex]

So, the next three terms of the series are:

[tex]T_2 = -4(\frac 13)^{2-1} = -\frac 43[/tex]

[tex]T_3 = -4(\frac 13)^{3-1} = -\frac 49[/tex]

[tex]T_4 = -4(\frac 13)^{4-1} = -\frac 4{27}[/tex]

So, the first four terms are -4, -4/3, -4/9 and -4/27

The common ratio of the series is 1/3 (1/3 is less than 1)

So, the series converges

The sum to infinity of the series is calculated as:

[tex]S_{\infty} = \frac{a}{1 -r}[/tex]

So, we have:

[tex]S_{\infty} = \frac{-4}{1 -1/3}[/tex]

[tex]S_{\infty} = \frac{-4}{2/3}[/tex]

Divide

[tex]S_{\infty} = -6[/tex]

Hence, the sum of the series is -6

Read more about geometric series at:

https://brainly.com/question/12006112

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