# F.5 Maths Qs About borrow loan

Peter borrows a loan of \$200000 from a bank at an interest rate of 6% per annum, compounded monthly. For each successive month after the day when the loan is

taken, loan interest is calculated and then a monthly instalment of \$x is

immediately paid to the bank until the loan is fully repaid (the last instalment may

be less \$x), where x<200000.

Prove that if Peter has not yet fully repaid the loan after paying the nth instalment, he still owes the bank \${200000(1.005)^(n) - 200x[(1.005)^(n) -1]}.

RE 邊位都好 & 知足常樂：

200000(1.005)^n - x(1.005)^(n-1) - x(1.005)^(n-2) - x(1.005)^(n-3) - ... - x

After paying the 2nd installment, he still owes

[200000(1.005) - x](1.005) - x

### 2 個解答

• 最愛解答

6% per annum = 0.5% per monthAfter paying the 1st installment, he still owes 200000(1.005) - x ;

After paying the 2nd installment, he still owes

[200000(1.005) - x](1.005) - x

= 200000(1.005)^2 - x(1.005) - xAfter paying the 3rd installment, he still owes[200000(1.005)^2 - x(1.005) - x](1.005) - x= 200000(1.005)^3 - x(1.005)^2 - x(1.005) - x⋯⋯So, after paying the nth installment, he still owes200000(1.005)^n - x(1.005)^(n-1) - x(1.005)^(n-2) - x(1.005)^(n-3) - ... - x= 200000(1.005)^n - [x + x(1.005) + x(1.005)^2 ... + x(1.005)^(n-1)]= 200000(1.005)^n - x[(1.005)^n - 1]/(1.005 - 1)= 200000(1.005)^n - 200x[(1.005)^n - 1]

Therefore, he still owes \${200000(1.005)^n - 200x[(1.005)^n - 1]}

2015-02-17 12:29:37 補充：

假設每個月還\$10000，則還第一次後尚欠

\$200000(1.005) - 10000 = \$191000

之後再計算利息，當然是基於\$191000，冇可能再基於\$200000，因為佢都已經冇欠咁多錢。

• In technical terms, you missed the "time value of money".

You pay \$x for n times, but the total is not worth \$nx.