Question

i need help answering 1-4! thanks

The mathematics of loans 99 Guided Project 45: The mathematics of loans Topics and skills: Sequences and sums Between houses, cars, or education, most people will take out a loan at some time in their lives. In a typical loan situation, a person borrows an amount $B at a fixed interest rate with a fixed payback period. The borrower makes monthly payments until the loan balance (the amount that remains to be paid) is reduced to zero. The goal of this project is to set up a mathematical model that describes the balance in the loan for cach month. The results of this project can be calculated and presented very effectively using a spreadsheet. 1. Let's take a specific example. Assume you borrow Bo - $15,000 with a fixed annual interest rate of 6%, or 0.5% per month. As a first problem, assume that your monthly payment is $500. The goal is to compute the number of months required to pay off the loan. Every month, two things happen: Interest, which is 0.5% of the current balance, is added to the current balance and the loan balance is decreased by the monthly payment of $500. We let B. be the loan balance after the nth payment. Explain why B. - Bo +0.005B0-500 - 1.005 Bo - 500 2. Explain why B. - 1.005B - 500 Show that in general, after the nth payment, the balance is B. - 1.005B, 500, for 1-1,2,3, 3. At this point we have found a recurrence relation for the sequence of loan balances. We now find an explicit formula for B. Beginning with the first month, we know B, -1.005 B – 500. After the second payment, the loan balance is B: -1.005B. - 500(I + 1.005). Show that after the third payment, the loan balance is B- 1.005 B-50001 + 1.005 +1.005) 4. Show that in general, the loan balance after the ath payment is B.- 1.005 B500(I + 1.005 +1.00 +...+1.005'), where - 01.2....

Answer 1

Every month two things happen:

1. Loan balance increases by the interest which is equal to 0.5% of current balance

2. $ 500 is paid by the borrower, so the balance reduces by $ 500

Part (1)

We start with B₀. At the end of 1st month or after first payment:

Loan balance, B₁ will be the net resultant of 1. & 2. above.

Hence, B₁ = B₀ + 0.5% x B₀ - 500 = B₀ + 0.005B₀ - 500 = 1.005B₀ - 500

Hence, B₁ = 1.005B₀ - 500

Part (2)

We start the next month (period) with starting balance of B₁. At the end of 2nd month or after second payment:

Loan balance, B₂ will be the net resultant of 1. & 2. above.

Hence, B₂ = B₁ + 0.5% x B₁ - 500 = B₁ + 0.005B₁ - 500 = 1.005B₁ - 500

Hence, B₂ = 1.005B₁ - 500

We can derive the recursive formula as:

B₁ = 1.005B₀ - 500

B₂ = 1.005B₁ - 500

.................

..............

B_n = 1.005B_n-1 - 500

Part (3)

B₁ = 1.005B₀ - 500

B₂ = 1.005B₁ - 500

Substitute B₁ from earlier equation in the second one to get,

B₂ = 1.005(1.005B₀ - 500) - 500 = 1.005²B₀ - 500(1 + 1.005)

SImilarly, B₃ = 1.005B₂ - 500 = 1.005(1.005²B₀ - 500(1 + 1.005)) - 500 = 1.005³B₀ - 500(1 + 1.005 + 1.005²)

Part (4)

Based on the observation below:

B₁ = 1.005¹B₀ - 500

B₂ = 1.005²B₀ - 500(1 + 1.005)

B₃ = 1.005³B₀ - 500(1 + 1.005 + 1.005²)

.................

................

B_n = 1.005ⁿB₀ - 500(1 + 1.005 + 1.005² +..............+ 1.005^n-1)