Lela invests $120,000 in a bank. The interest rate is 4.65% per annum, compounded monthly. a) Write a recurrence relation to model Lela's investment. (Note that the question refers to Mikayla, but should be Lela). b) Determine the amount of money in Lela's account after 2 years.

1. Problem Description

Lela invests $120,000 in a bank. The interest rate is 4.65% per annum, compounded monthly.

a) Write a recurrence relation to model Lela's investment. (Note that the question refers to Mikayla, but should be Lela).

b) Determine the amount of money in Lela's account after 2 years.

2. Solution Steps

a) Recurrence relation:

Let $A_n$ be the amount in the account after $n$ months.

The initial amount is $A_0 = 120000$.

The annual interest rate is 4.65%, so the monthly interest rate is $\frac{4.65\%}{12} = \frac{0.0465}{12} = 0.003875$.

Each month, the amount in the account is multiplied by $1 + 0.003875 = 1.003875$.

Therefore, the recurrence relation is:

$A_{n+1} = A_n \times (1 + \frac{0.0465}{12}) = A_n \times 1.003875$

$A_0 = 120000$

b) Amount after 2 years:

Since the interest is compounded monthly, we need to find the amount after $2 \times 12 = 24$ months.

The formula for compound interest is:

$A = P(1 + r/n)^{nt}$

where:

$A$ = the future value of the investment/loan, including interest

$P$ = the principal investment amount ($120000)

$r$ = the annual interest rate (decimal) ($0.0465)

$n$ = the number of times that interest is compounded per year ($12)

$t$ = the number of years the money is invested or borrowed for ($2)

$A = 120000(1 + \frac{0.0465}{12})^{12 \times 2}$

$A = 120000(1 + 0.003875)^{24}$

$A = 120000(1.003875)^{24}$

$A = 120000 \times 1.096353$

$A = 131562.36$

3. Final Answer

a) The recurrence relation is:

$A_{n+1} = 1.003875 A_n$

$A_0 = 120000$

b) The amount of money in Lela's account after 2 years is approximately $131562.36.