Answer
$a) A_n = (0.8)^n (4,500,000,000)$
b) ${A_1 = $3,600,000,000}$
${A_2 = $2,880,000,000}$
${A_3 = $2,304,000,000}$
${A_4 = $1,843,200,000}$
$c) \lim\limits_{n \to \infty} A_n = 0$; therefore, $ A_n $ converges
Work Step by Step
a) Because the balance is 80% of the previous balance at each value of n, r = 0.8. The initial balance is 4,500,000,000 dollars. Therefore, ${A_n = (0.8)^n (4,500,000,000)} $
b) ${A_1 = (0.8)(4,500,000,000) = $3,600,000,000}$
${A_2 = (0.8)^{2}(4,500,000,000) = $2,880,000,000}$
${A_3 = (0.8)^{3}(4,500,000,000) = $2,304,000,000}$
${A_4 = (0.8)^{4}(4,500,000,000) = $1,843,200,000}$
c) $\lim\limits_{n \to \infty} A_n $
$= \lim\limits_{n \to \infty} (0.8)^{n}(4.5)$
$= (0.8)^{\infty} (4.5)$
$= (0)(4.5)$
$= 0$
Because $\lim\limits_{n \to \infty} A_n = 0,$ the series converges.
