Deriving the Future Value of an Ordinary Annuity

Why Payment Timing Determines Future Value

A series of equal deposits does not earn interest for the same length of time. In an ordinary annuity, each payment is made at the end of a period. When the value is measured at the date of the last payment, the first deposit has accumulated for $n-1$ periods, the second for $n-2$, and the last deposit for no time at all. The future value is the sum of those separately accumulated amounts.

That timing is the central idea behind the ordinary-annuity accumulation formula. The formula is not a separate rule to memorize without context: it follows directly from moving each payment to one common valuation date and adding the resulting geometric series.

Definition: An ordinary annuity is a sequence of $n$ equal payments, each made at the end of one successive period. If each payment is $R$, the interest rate per payment period is $i$, and the value is measured immediately after the $n$th payment, the accumulated value is $R s_{\overline{n}|}$. For $i\ne 0$, $s_{\overline{n}|}=\frac{(1+i)^n-1}{i}$; for $i=0$, $s_{\overline{n}|}=n$.

Here $n$ is a positive integer and $i$ is the effective rate for one payment period. The usual accumulation model assumes that the same rate applies in every period, that interest is compounded once per payment period, and that each payment is made in full at the stated time. For a positive accumulation factor, assume $i>-1$. Rates can be annual, quarterly, monthly, or another period rate, but the rate and payment period must match.

Deriving the Accumulated-Value Formula

Place the valuation date immediately after the final payment. The payment at the end of period 1 earns interest for $n-1$ periods; the payment at the end of period 2 earns interest for $n-2$ periods. Continuing this pattern, the final payment earns interest for zero periods. Under a constant effective rate $i$, a payment carried forward $k$ periods is multiplied by $(1+i)^k$.

\begin{aligned} \text{Value at time }n &=R(1+i)^{n-1}+R(1+i)^{n-2}+\cdots+R(1+i)+R\\ &=R\left[1+(1+i)+(1+i)^2+\cdots+(1+i)^{n-1}\right]. \end{aligned}

The bracketed terms form a finite geometric series with first term 1, common ratio $1+i$, and $n$ terms. To justify its sum, let $G=1+(1+i)+\cdots +(1+i)^{n-1}$. Multiplying by $1+i$ shifts each power up by one. Subtracting the original series cancels every intermediate power:

\begin{aligned} (1+i)G&=(1+i)+(1+i)^2+\cdots +(1+i)^n,\\ (1+i)G-G&=(1+i)^n-1,\\ iG&=(1+i)^n-1. \end{aligned}

When $i\ne 0$, division by $i$ gives $G=\frac{(1+i)^n-1}{i}$. Multiplying by $R$ gives the future value. When $i=0$, each payment remains unchanged, so the value is $nR$; this is also the limit of the formula as $i$ approaches zero. Thus the formula follows from the cash flows and their accumulation periods, rather than from a timing convention hidden in the notation.

\boxed{ R s_{\overline{n}|} =R\left[1+(1+i)+\cdots +(1+i)^{n-1}\right] =R\frac{(1+i)^n-1}{i},\quad i\ne 0 }

A useful timing check is to inspect the exponents: they must be $n-1,n-2,\ldots,1,0$ when the value is measured at the date of the last payment. There are $n$ terms, and the final payment contributes $R$, not $R(1+i)$, because it has not earned a period of interest at that valuation date.

Worked Applications

Worked Example: Six Annual Deposits at a Positive Rate

A hypothetical account receives deposits of $900 at the end of each year for 6 years. The effective annual interest rate is $i=0.03$, credited annually. Find the account value immediately after the sixth deposit, and separate the accumulated interest from the deposits. This is a deterministic calculation; no mortality assumption or probability distribution is involved.

Method and timing. Use the ordinary-annuity future-value formula with $R=900$, $n=6$, and $i=0.03$. At the end of year 6, the deposits have accumulated for 5, 4, 3, 2, 1, and 0 years, respectively. The last deposit therefore is not multiplied by $1.03$.

\begin{aligned} s_{\overline{6}|} &=\frac{(1.03)^6-1}{0.03}\\ &=\frac{1.194052296529-1}{0.03}\\ &=6.4684098843,\\ \text{Accumulated value} &=900(6.4684098843)\\ &=5821.56889587. \end{aligned}

The total deposits are $6(900)=5400$. Thus the accumulated interest is $5821.56889587-5400=421.56889587$. Rounded to the nearest cent, the account value is $5,821.57 and accumulated interest is $421.57. The check is sensible: the account value exceeds the deposits because the rate is positive, while the final $900 deposit earns no interest by the valuation date.

Common mistake avoided: Treating all six deposits as though they earned six full years would multiply each by $(1.03)^6$. That would value cash flows at the wrong dates. The exponents in the correct calculation descend from 5 to 0.

Worked Example: Quarterly Payments and a Nominal Rate

A hypothetical reserve account receives $500 at the end of each quarter for 8 quarters. The stated rate is a nominal annual rate of $i^{(4)}=0.06$, convertible quarterly. Find the value immediately after the eighth payment. There is no mortality contingency or random loss; the cash flows and rate are deterministic.

Method and rate conversion. Because payments are quarterly, use the rate per quarter. A nominal rate of 6% convertible quarterly gives $i^{(4)}/4=0.06/4=0.015$, or 1.5% per quarter. There are $n=8$ payments and $R=500$.

\begin{aligned} (1.015)^8&=1.1264925865953,\\ s_{\overline{8}|} &=\frac{1.1264925865953-1}{0.015}\\ &=8.43283910635,\\ \text{Accumulated value} &=500(8.43283910635)\\ &=4216.419553175. \end{aligned}

The deposits total $8(500)=4000$, so accumulated interest is $4216.419553175-4000=216.419553175$. Rounded to the nearest cent, the value is $4,216.42 and the interest is $216.42. Since the rate per quarter is positive, the value is greater than the deposits; the last $500 is included without quarterly interest.

Common mistake avoided: Substituting 0.06 as the quarterly rate would apply a 6% rate every three months, not the stated 6% nominal annual rate convertible quarterly. The required per-period rate is 0.015.

Worked Example: A Negative Rate and a Direct Sum Check

A hypothetical account receives $1,200 at the end of each year for 5 years. Its effective annual rate is $i=-0.02$, credited annually. Find the value immediately after the fifth deposit. No mortality assumption or probability distribution applies.

Method. Use $R=1200$, $n=5$, and $i=-0.02$. The accumulation factor per year is $1+i=0.98$, so earlier deposits decrease in value under this negative-rate assumption.

\begin{aligned} s_{\overline{5}|} &=\frac{(0.98)^5-1}{-0.02}\\ &=\frac{0.9039207968-1}{-0.02}\\ &=4.80396016,\\ \text{Accumulated value} &=1200(4.80396016)\\ &=5764.752192. \end{aligned}

A direct cash-flow check gives the same factor: $1+0.98+0.9604+0.941192+0.92236816=4.80396016$. The deposits total $5(1200)=6000$, so the accumulated interest is $5764.752192-6000=-235.247808$. Rounded to the nearest cent, the value is $5,764.75; the account has lost $235.25 relative to the deposits. The result is below $6,000, as it should be for this negative rate.

Common mistake avoided: A negative rate does not mean that the annuity formula is unusable. Here $i\ne0$ and $1+i=0.98>0$, so the geometric sum and formula both apply. The negative numerator and negative denominator in the factor produce a positive accumulation factor.

Interpreting the Formula and Checking Results

The accumulation factor $s_{\overline{n}|}$ is the future value of one unit paid at the end of each period, under the stated rate and timing. Multiplying it by $R$ scales that unit-payment result to the actual payment amount. The factor is not a present value: it describes the value at the final payment date. Its $n$ terms reflect the $n$ actual payments, including one payment with zero accumulation periods.

For $i>0$, each earlier payment grows, so $R s_{\overline{n}|}$ exceeds the total deposits $nR$. For $i=0$, it equals $nR$. For $-1<i<0$, earlier payments shrink, so the accumulated value is less than $nR$, assuming positive deposits. These comparisons are useful checks, not replacements for a timeline.

A practical sequence for solving an ordinary-annuity future-value problem is:

Set the valuation date.
For this formula, measure the value immediately after the last end-of-period payment.

Match periods.
Identify the number of payments and express the interest rate per payment period. Convert a nominal rate if necessary.

Accumulate and sum.
Use $R s_{\overline{n}|}$, or write the individual accumulated payments to verify the exponents.

Check the result.
Compare with total deposits $nR$, verify the last payment earns no interest at the chosen date, and round money only at the end.

The formula does not directly apply when payment amounts differ, the rate changes by period, or the valuation date is not the date immediately after the final payment. In those cases, the safe starting point remains the same: carry each cash flow to the requested date using the rate applicable over its accumulation period, then add the values. The compact annuity factor relies on level payments, a constant per-period rate, and the specified ordinary-annuity timing.

Common Exam Traps

Using the wrong number of interest periods. With $n$ end-of-period payments valued at the last payment date, the first payment earns $n-1$ periods, not $n$.
Mixing annual and payment-period rates. A quarterly annuity needs a quarterly rate in the formula. Convert a nominal annual rate to the rate per quarter before substituting.
Misreading the valuation date. If the requested value is later than the final payment date, first find the annuity value at the final payment date, then accumulate that value for the additional periods.
Dividing by zero when $i=0$. At zero interest, each payment remains $R$, so the future value is $nR$.
Rounding too early. Keep sufficient precision in $(1+i)^n$ and the annuity factor, then round the final dollar amount to cents.

Key takeaway: The ordinary-annuity future value is obtained by accumulating every end-of-period payment to the date immediately after the last payment. The resulting powers run from $n-1$ down to 0, and their geometric sum gives $s_{\overline{n}|}=\frac{(1+i)^n-1}{i}$ for $i\ne0$. This derivation and its timing checks are aligned with the SOA VEE Accounting and Finance syllabus.

Check Your Understanding

For each problem, use the stated payment timing and rate period. Answers are rounded to the nearest cent where money is requested.

A maintenance fund receives $250 at each year-end for 4 years and earns an effective annual rate of 4%, compounded annually. Find its value immediately after the fourth deposit. (Answer: $1,061.62)
A hypothetical account receives $700 at the end of each year for 6 years at an effective annual rate of 2.5%. Find its accumulated value and the accumulated interest. (Answer: value $4,471.42; interest $271.42)
A monthly reserve receives $300 at each month-end for 10 months. The nominal annual rate is 4.8% convertible monthly. Find the value immediately after the tenth deposit. (Answer: $3,054.58)
A fund receives $1,000 at each year-end for 5 years, with an effective annual rate of -1%. Find the accumulated value immediately after the fifth payment and compare it with total deposits. (Answer: $4,901.00, which is $99.00 below the $5,000 deposited)
Four deposits of $1,500 are made at the ends of years 1 through 4 in an account earning 5% effective annually. Find the value at the end of year 6. (Answer: $7,127.87)

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