Answer
a) a_n = 1.09 ⋅ an−1
b) a_n = 1000 ⋅ 1.09^n
c) $5529041
Work Step by Step
A.
a_0 = 1000; r = 0.09 (9%)
Each year, 9% interest is added to account considering that ‘a_n’ is the amount in the account at the end of ‘n’ years, resulting in the amount being equal to the previous year’s amount plus 9% of the previous year’s amount. therefore,
a_n = a_n-1 + 0.09 ⋅ a_n-1
a_n = 1.09 ⋅ a_n-1
B.
a_n = 1.09 ⋅ an−1; a_0 = 1000
Apply the recurrence relation on this problem:
a_n = 1.09a_n-1 = 1.09^1a_n−1
a_n = 1.09(1.09a^n-2) = 1.09^2a_n−2
a_n = 1.09^2(1.09a^n-3) = 1.09^3a_n−3
a_n = 1.09^3(1.09a^n-4) = 1.09^4a_n−4
……..
a_n = 1.09^a_n -n
a_n =1.09^n a_0
a_n = 1000 ⋅ 1.09^n
C.
We use the result in ‘b’ and evaluate n = 100
a_100 = 1000 1.09^100 = $5529041 (account after 100 years)
