A Closed-Form Balance After Any Number of Payments

Why a Closed-Form Balance Is Useful

An amortization schedule gives the balance after each payment, but a question may ask for the balance after a payment far into the term. Listing every row is slow and creates opportunities for arithmetic errors. A closed-form formula gets that balance directly from the original principal, the regular payment, the periodic interest rate, and the number of payments made.

The timing convention matters. In this tutorial, payments are made at the end of each payment period, the interest rate is constant from period to period, and each payment is the same amount. The balance $B_k$ means the amount owed immediately after the $k$th payment. Thus $B_0$ is the original principal, before any payments have been made.

Definition: For a loan with original principal $L$, level end-of-period payment $R$, and effective interest rate $i$ per payment period, $B_k$ is the outstanding balance immediately after $k$ payments. The rate $i$ must apply to the interval between payments. The closed-form balance is $$ B_k=L(1+i)^k-Rs_{\overline{k}|}, $$ where, for $i\ne 0$, $$ s_{\overline{k}|}=\frac{(1+i)^k-1}{i}. $$ For $i=0$, the balance is $B_k=L-kR$.

The formula applies while the stated level payments are being made under the same rate and timing assumptions. It can also be used when the payment does not fully amortize the loan by a specified date; in that case, the formula gives the balance after those payments, not a claim that the loan is fully repaid. If the actual final payment is adjusted, use the regular-payment formula only through the payments that really equal $R$.

Deriving the Formula From the Balance Recurrence

In one period, the balance earns interest at rate $i$, then the end-of-period payment $R$ is subtracted. Therefore, the recurrence is $B_k=(1+i)B_{k-1}-R$, with $B_0=L$. The closed form follows by repeatedly applying that recurrence.

Accumulate the previous balance.
After $k$ periods, the original principal contributes $L(1+i)^k$, before accounting for payments.

Accumulate each payment to time $k$.
The first payment earns interest for $k-1$ periods, the second for $k-2$, and the $k$th payment earns none. Their accumulated total is $R[(1+i)^{k-1}+(1+i)^{k-2}+\cdots+1]$.

Subtract the accumulated payments.
The powers form a geometric series, whose sum is $s_{\overline{k}|}$. Subtracting that sum of accumulated payments from the accumulated original principal gives $B_k=L(1+i)^k-Rs_{\overline{k}|}$.

The geometric-series step explains why the factor is $s_{\overline{k}|}$, not $a_{\overline{k}|}$. The payments are accumulated forward to the time of the balance calculation. When $i=0$, each of the $k$ payments remains $R$ at that time, so the same argument gives $B_k=L-kR$.

\begin{aligned} B_k &=L(1+i)^k-R\sum_{j=0}^{k-1}(1+i)^j\\ &=L(1+i)^k-R\frac{(1+i)^k-1}{i}, \qquad i\ne 0. \end{aligned}

This is a forward accumulation calculation: first bring the original principal to the date immediately after payment $k$, then subtract the value at that same date of all $k$ payments. Both terms must be measured at the same time point.

Using the Formula in Loan Problems

Before substituting, identify the payment interval and determine the effective rate for that interval. If the quoted rate is nominal, convert it to the payment period. For a nominal annual rate $i^{(m)}$ convertible $m$ times per year, the rate per conversion period is $i^{(m)}/m$. If payments occur each conversion period, that is the $i$ used in the balance formula. If payment and conversion frequencies differ, first find the effective rate over the actual payment interval.

Then establish the payment count and the balance timing. “After six payments” means use $k=6$, with the sixth payment already subtracted. A balance just before a payment is a different quantity: the current balance has earned interest for the period, but that payment has not yet been subtracted.

Worked Example: Balance After Three Annual Payments

A hypothetical loan has an original principal of $8,000. Interest is 5% effective per year, and the borrower makes end-of-year payments of $2,000. Find the balance immediately after the third payment.

Method. Use the closed-form balance with $L=8000$, $i=0.05$, $R=2000$, and $k=3$. The payment and interest periods are both annual, so no rate conversion is needed.

First calculate the accumulated original principal: $$ 8000(1.05)^3=8000(1.157625)=9261. $$ Next calculate the accumulated-value annuity factor: $$ s_{\overline{3}|}=\frac{(1.05)^3-1}{0.05} =\frac{1.157625-1}{0.05} =3.1525. $$ The three payments accumulated to immediately after payment 3 are: $$ 2000(3.1525)=6305. $$ Subtract: $$ B_3=9261-6305=2956. $$ The balance immediately after the third payment is $2,956.00.

Check. Applying the recurrence gives $B_1=8000(1.05)-2000=6400$, $B_2=6400(1.05)-2000=4720$, and $B_3=4720(1.05)-2000=2956$, matching the formula. The common mistake to avoid is using a discounted annuity factor for payments that must be accumulated to the balance date.

Worked Example: Convert a Nominal Rate Before Finding the Balance

A hypothetical loan of $20,000 charges a nominal annual rate of 12% convertible monthly. The borrower pays $1,200 at the end of each month. Find the balance immediately after the sixth payment.

Method. Because interest is convertible monthly and payments are monthly, the periodic rate is $$ i=\frac{0.12}{12}=0.01. $$ Use $L=20000$, $R=1200$, $i=0.01$, and $k=6$.

The original principal accumulated for six months is $$ 20000(1.01)^6=20000(1.061520150601)=21230.40301202. $$ The annuity factor is $$ s_{\overline{6}|}=\frac{1.061520150601-1}{0.01} =6.1520150601. $$ The accumulated value of the six payments is $$ 1200(6.1520150601)=7382.41807212. $$ Therefore, $$ B_6=21230.40301202-7382.41807212 =13847.98493990. $$ Rounded to the nearest cent, the balance is $13,847.98.

Check. The formula gives $B_5=14898.99499$ (to the displayed precision), so one more month of interest and the sixth payment give $14898.99499(1.01)-1200=13847.98494$, consistent with the result. The common mistake is to use 12% as the monthly rate; the rate per month is 1%, not 12%.

Worked Example: Quarterly Payments and a Nominal Quarterly Rate

A hypothetical $30,000 loan has a nominal annual rate of 8% convertible quarterly. Payments of $1,800 are made at the end of each quarter. Find the balance immediately after the fifth payment.

Method. There are four conversion periods per year, and payments are quarterly, so the periodic rate is $$ i=\frac{0.08}{4}=0.02. $$ Use $L=30000$, $R=1800$, and $k=5$.

Accumulate the original principal: $$ 30000(1.02)^5=30000(1.1040808032)=33122.424096. $$ Calculate the annuity factor: $$ s_{\overline{5}|}=\frac{1.1040808032-1}{0.02} =5.20404016. $$ Accumulate the five payments: $$ 1800(5.20404016)=9367.272288. $$ Thus, $$ B_5=33122.424096-9367.272288=23755.151808. $$ Rounded to the nearest cent, the balance is $23,755.15.

Check. The formula gives $B_4=25054.0704$. One more quarter of interest followed by the fifth payment gives $25054.0704(1.02)-1800=23755.151808$, which agrees. The key trap is mixing annual and quarterly units: both the payment count $k$ and the rate $i$ here are quarterly.

Common Errors and Exam Traps

The main safeguard is to keep the balance date, payment count, and rate period aligned. Each payment included in $s_{\overline{k}|}$ must be one of the $k$ payments already made, and the formula returns a balance immediately after the last of them.

Using the wrong annuity factor. The payments are accumulated to the balance date, so use $s_{\overline{k}|}$. A present-value factor discounts payments and does not represent their accumulated value at that date.
Using the annual rate as the periodic rate. Convert a nominal rate to the payment interval before substitution. A monthly payment count requires a monthly periodic rate.
Shifting the payment count. After $k$ payments, use $k$, not $k-1$. The $k$th payment is included and is not accumulated for any additional period.
Ignoring the zero-rate case. The quotient defining $s_{\overline{k}|}$ divides by $i$, so it should not be used at $i=0$. Use $B_k=L-kR$ instead.
Assuming any quoted payment guarantees full repayment. The formula calculates the balance from the stated payment and rate. Whether a contract fully amortizes, or adjusts its final payment, depends on its terms.

Key takeaway: For $k$ level payments made at period-end, accumulate the original principal for $k$ periods and subtract the accumulated value of the $k$ payments: $B_k=L(1+i)^k-Rs_{\overline{k}|}$. Use the effective rate per payment period and interpret the result immediately after payment $k$.

Check Your Understanding

For each problem, identify the periodic rate and payment count before applying the balance formula.

A hypothetical loan has principal $9,600, no interest, and end-of-period payments of $800. Find the balance immediately after seven payments. (Answer: $4,000.00)
A hypothetical loan of $50,000 charges a nominal annual rate of 6% convertible quarterly. Quarterly payments are $3,500. Find the balance immediately after eight payments. (Answer: $26,809.69)
A hypothetical loan has principal $18,000, an effective annual rate of 4%, and annual end-of-year payments of $4,000. Find the balance immediately after three payments. (Answer: $7,761.15)
A hypothetical loan of $25,000 earns interest at 10% effective annually. Payments of $6,000 are made at each year-end. Find the balance immediately after three payments. (Answer: $13,415.00)

What You'll Learn