Question

PLEASE SHOW ALL WORK.

Answer 1

Answer:

a. T1 = -4 , T2 = -4/3 , T3 = -4/9 , T4 = -4/27

b. It is converge

c. The sum to ∞ = -6

Step-by-step explanation:

a.

∵ Tn = -4(1/3)^(n-1)

∵ The lower n = 1

∵ The geometric series Tn = a(r)^(n-1)

∴ T1 = a , T2 = ar , T3 = ar² , T4 = ar³

∴ a = -4 , r = 1/3

∴ T1 = -4

∴ T2 = -4(1/3) = -4/3

∴ T3 = -4(1/3)² = -4/9

∴ T4 = -4(1/3)³ = -4/27

b.

r = 1/3

∴ -1 < r < 1

∴ It is converge because the value of 1/3 when n is a very large

  number will approach to zero

c.

∵ The sum of the geometric series = a(1-(r)^n)/1-r

∵ r^n ≅ 0 when n is a very large number

∴ The sum to ∞ = a/1 - r = -4/(1 - 1/3) = -4/(2/3) = -6