Deriving the Accumulated Value of an Annuity-Immediate

Why Accumulated Value Depends on Payment Timing

A sequence of equal deposits does not all earn interest for the same length of time. In an annuity-immediate, each payment is made at the end of a period. The first payment earns interest for almost the entire term, while the final payment is made at the valuation date and earns no interest before that date. To find the accumulated value, carry each payment forward to a common date and add the resulting amounts.

The derivation below uses compound interest at a constant effective rate per payment period. It focuses on the value immediately after the final payment. This timing is the key distinction between the accumulated value of an annuity-immediate and other annuity values.

Definition: An annuity-immediate with $n$ payments of $R$ is a series of equal payments made at the ends of $n$ consecutive periods. If the effective interest rate per period is $i$, its accumulated value at the end of period $n$, immediately after the last payment, is denoted $R s_{\overline{n}|}$. The factor $s_{\overline{n}|}$ is the accumulated-value factor for the annuity.

Assume $n$ is a positive integer, the rate $i$ is constant from period to period, and each payment is made as scheduled. Payments earn compound interest from their payment dates to the valuation date. The rate must be expressed per payment period: if payments are monthly, for example, the rate used in the formula must be an effective monthly rate. The model excludes fees, taxes, withdrawals, and changes in the interest rate.

Deriving the Accumulated-Value Factor

Place the payments at times $1,2,\ldots,n$, and value the fund at time $n$, immediately after payment $n$. A payment made at time $k$ has $n-k$ complete periods to earn interest. Its value at time $n$ is therefore $R(1+i)^{n-k}$. Listing the payments from earliest to latest gives the accumulated amounts $R(1+i)^{n-1}, R(1+i)^{n-2}, \ldots, R(1+i), R$.

Adding these amounts and factoring out $R$ gives a finite geometric series. The factor is written with the standard actuarial notation $s_{\overline{n}|}$.

R s_{\overline{n}|} =R\left[(1+i)^{n-1}+(1+i)^{n-2}+\cdots+(1+i)+1\right]

For $i\ne 0$, set $q=1+i$. If $S=1+q+q^2+\cdots+q^{n-1}$, then multiplying by $q$ and subtracting the original sum cancels all interior terms:

qS-S=(q+q^2+\cdots+q^n)-(1+q+\cdots+q^{n-1})=q^n-1

Thus $(q-1)S=q^n-1$. Since $q-1=i$, the factor is $\big((1+i)^n-1\big)/i$. If $i=0$, no payment earns interest, so the accumulated value is simply $nR$. The quotient formula is not used at $i=0$, because it would require division by zero.

Theorem: For $n$ equal end-of-period payments of $R$, accumulated immediately after the $n$th payment at a constant effective rate $i$ per period, $$ s_{\overline{n}|}= \begin{cases} \dfrac{(1+i)^n-1}{i}, & i\ne 0,\\[6pt] n, & i=0. \end{cases} $$ The accumulated value is $R s_{\overline{n}|}$.

Mark the payment dates.
For an annuity-immediate, payments occur at the ends of periods $1$ through $n$.

Choose the focal date.
For $R s_{\overline{n}|}$, use the date immediately after payment $n$.

Count each payment’s interest periods.
The payment at time $k$ earns interest for $n-k$ periods.

Add the accumulated payments.
Use the geometric-series factor, then multiply by $R$.

Worked Applications

In each application, the number of compounding periods must match the number of payment periods. The formula is a compact way to add the individually accumulated payments; it does not change the payment dates.

Worked Example: Five Annual Deposits to a Reserve

A hypothetical reserve receives deposits of $600 at the end of each year for five years. The account earns an effective annual interest rate of 5%. Find the accumulated value immediately after the fifth deposit.

Method. There are five end-of-year payments, and the focal date is the end of year 5. The first deposit earns interest for four years, the next for three, then two, then one; the fifth earns none before the focal date.

\begin{aligned} \text{Accumulated value} &=600\left[(1.05)^4+(1.05)^3+(1.05)^2+(1.05)+1\right]\\ &=600\left[1.21550625+1.157625+1.1025+1.05+1\right]\\ &=600(5.52563125)\\ &=3315.37875 \end{aligned}

Equivalently, $s_{\overline{5}|}=\big((1.05)^5-1\big)/0.05=(1.2762815625-1)/0.05=5.52563125$. Multiplying by $600 gives $3,315.37875, or $3,315.38 to the nearest cent.

Check. The final deposit is exactly $600 at the valuation date, and each of the other four deposits accumulates to more than $600. The result, $3,315.38, is therefore greater than the total deposits of $3,000, as expected at a positive interest rate.

Worked Example: Semiannual Contributions at a Nominal Rate

A hypothetical equipment fund receives eight deposits of $400, one at the end of each half-year. The account earns a nominal annual interest rate of 6% convertible semiannually. Find the value immediately after the eighth deposit.

Method. A nominal rate convertible twice per year gives an effective rate of $0.06/2=0.03$ per half-year. The eight deposits span eight half-year periods, so use $n=8$ and $i=0.03$, not the annual rate in the annuity factor.

\begin{aligned} s_{\overline{8}|} &=\frac{(1.03)^8-1}{0.03}\\ &=\frac{1.266770081387616-1}{0.03}\\ &=8.89233604625387 \end{aligned}

The accumulated value is

$$ 400(8.89233604625387)=3556.934418501548 $$

Thus the fund is worth $3,556.93 to the nearest cent. As a check on the timing, the eighth deposit contributes $400, and the first contributes $400(1.03)^7$; the eight contributions together are represented by the factor just calculated.

Worked Example: Deposits Stop Before the Valuation Date

A hypothetical training grant account receives six annual deposits of $500 at the ends of years 1 through 6. It earns an effective annual rate of 5%. Find its value at the end of year 8, assuming no further deposits are made.

Method. First accumulate the six deposits to the end of year 6, immediately after the final deposit. Then accumulate that balance for two more years. The annuity factor applies to the six payment periods, not to all eight years.

\begin{aligned} s_{\overline{6}|} &=\frac{(1.05)^6-1}{0.05}\\ &=\frac{1.340095640625-1}{0.05}\\ &=6.8019128125 \end{aligned}

The value at the end of year 6 is $500(6.8019128125)=3400.95640625$. This balance earns interest for two additional years:

\begin{aligned} \text{Value at year 8} &=3400.95640625(1.05)^2\\ &=3400.95640625(1.1025)\\ &=3749.554437890625 \end{aligned}

The account value at the end of year 8 is $3,749.55 to the nearest cent. The total deposits are $3,000, so a value above $3,000 is consistent with earning interest. The two extra years of accumulation apply to the year-6 balance; they do not represent two extra deposits.

Common Timing and Formula Errors

The most common error is to give every payment one more period of interest than it actually earns. In the first example, using $600(1.05)s_{\overline{5}|}$ would give $600(1.05)(5.52563125)=3481.1476875$, or $3,481.15. That is one year too much growth: it treats even the fifth deposit as if it had been made a year before the valuation date. The correct value is $600s_{\overline{5}|}=$3,315.38$, because the fifth deposit is made at the focal date and earns zero periods of interest before that date.

Another error is to use $n$ periods of interest for the first deposit and then count $n$ payments as well. For $n$ end-of-period payments valued at time $n$, the exponents run from $n-1$ down to $0$, not from $n$ down to $1$. A quick timeline prevents this off-by-one error.

Using the wrong rate period: If the payment interval is a half-year, the factor needs the effective rate per half-year.
Confusing total deposits with accumulated value: Total deposits are $nR$; accumulated value also reflects interest earned by earlier deposits.
Using the quotient at zero interest: When $i=0$, use $nR$ directly.
Ignoring the focal date: If valuation occurs after the last deposit, first value the annuity at its last payment date and then accumulate that balance for the additional periods.

The result is aligned with the SOA Exam FM (Financial Mathematics) syllabus because it requires candidates to connect payment timing, interest periods, and a focal date rather than simply insert values into a formula.

Key takeaway: An annuity-immediate accumulates by carrying each payment forward to the date immediately after the final payment. The first payment earns $n-1$ periods of interest and the last earns none, producing $s_{\overline{n}|}=\big((1+i)^n-1\big)/i$ when $i\ne0$.

Check Your Understanding

For each problem, identify the payment period and valuation date before applying the factor. Round monetary answers to the nearest cent.

A savings account receives three deposits of $100 at the ends of years 1, 2, and 3. The effective annual rate is 10%. Find the value immediately after the third deposit. (Answer: $331.00)
A laboratory reserve receives four deposits of $250 at the end of each year. The effective annual rate is 4%. Find its value immediately after the fourth deposit. (Answer: $1,061.62)
A hypothetical fund receives six deposits of $180 at the ends of successive half-years. The nominal annual rate is 8% convertible semiannually. Find its value immediately after the sixth deposit. (Answer: $1,193.94)
A training account receives four annual deposits of $700 at the ends of years 1 through 4. The effective annual rate is 3%, and no further deposits are made. Find the value at the end of year 6. (Answer: $3,106.89)
An account receives nine end-of-year deposits of $125 and earns no interest. Find its value immediately after the ninth deposit. (Answer: $1,125.00)

What You'll Learn