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.

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∞ = ∞

answer is 144

step-by-step explanation:

a; x = 0.5

step-by-step explanation:

-10x = -5

x = 0.5