Guy invests $S c$ at the beginning of each quarter into an account earming interest at a rate of $4 %$ per annum, compounding quarterly, where $c$ is your assigned value in the table above.
The amount in the account immediately after the $n^{th}$ quarter can be determined using the recurrence relation
$A_{n} = A_{n - 1} (1.01) + c (1.01)$
where $n = 1, 2, 3, \dots$ and $A_{0} = 0$
(a) Use the recurrence relation to find the amount of money in the account at the end of
2 the third quarter.
(b) Calculate the amount of interest earned in the first three quarters.
1
(c) Calculate the amount of money in the account at the end of 10 years.
2

Answer

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Answer

\boxed{A_{40} = c(1.01)(1.01^{40} - 1)/(1.01 - 1)}

Steps

Step 1 ：A_1 = A_0(1.01) + c(1.01) = 0(1.01) + c(1.01) = c(1.01)

Step 2 ：A_2 = A_1(1.01) + c(1.01) = c(1.01)(1.01) + c(1.01) = c(1.01)^2 + c(1.01)

Step 3 ：A_3 = A_2(1.01) + c(1.01) = (c(1.01)^2 + c(1.01))(1.01) + c(1.01) = c(1.01)^3 + c(1.01)^2 + c(1.01)

Step 4 ：\boxed{A_3 = c(1.01)^3 + c(1.01)^2 + c(1.01)}

Step 5 ：I_3 = A_3 - 3c = c(1.01)^3 + c(1.01)^2 + c(1.01) - 3c

Step 6 ：\boxed{I_3 = c(1.01)^3 + c(1.01)^2 + c(1.01) - 3c}

Step 7 ：A_{40} = c(1.01)^{40} + c(1.01)^{39} + \cdots + c(1.01)^2 + c(1.01)

Step 8 ：\boxed{A_{40} = c(1.01)(1.01^{40} - 1)/(1.01 - 1)}