Compound interest as repeated SI

Every compound interest sum a Class 9 student meets can, in principle, be solved without ever writing the "famous" CI formula — by doing simple interest again and again, one year at a time, on a principal that keeps growing. That is the idea behind compound interest as repeated SI: find the interest for Year 1 on the starting principal, add it in, then find the interest for Year 2 on this new, larger amount, and so on. This year-by-year build-up is not a shortcut to skip — it is why the compact formula $A = P(1+r/100)^n$ works at all, and it is the only method available when the rate of interest changes from year to year, a favourite ICSE examiner twist. This article rebuilds the formula from repeated SI and drills the technique on both a constant-rate and a changing-rate example.

The core insight

Compound interest is simple interest recalculated every year on a principal that has already absorbed the previous year's interest, so repeating the SI formula year by year — and multiplying the growth factor $(1+r/100)$ once per year — is both the meaning of "compounding" and the derivation of the CI formula itself.

The core idea

Simple interest keeps the principal frozen. In SI, the amount $P$ used in the interest calculation never changes, no matter how many years pass:

SI = \dfrac{P \times r \times t}{100}

where: $P$ is the original principal, $r$ is the rate per cent per annum, and $t$ is the time in years. Every year earns exactly the same interest, because $P$ is the same input every time.

Compound interest lets the principal grow. After each year, that year's interest is folded into the principal, so the next year's SI is computed on a bigger number. This single change — recycling the amount, not the original principal, as the base for the next year's interest — is the entire difference between SI and CI, and it is why CI is described as "interest on interest."

The repeated-SI recipe for $n$ years. Compute $SI_1$ on $P_1 = P$ ; form the new principal $P_2 = P_1 + SI_1$ (the amount at the end of Year 1); compute $SI_2$ on $P_2$ ; form $P_3 = P_2 + SI_2$ ; and continue until Year $n$ . The total compound interest is the sum $SI_1 + SI_2 + \cdots + SI_n$ , and the final amount is $A = P + \text{total CI}$ . Because each year's SI is calculated on a principal that already includes every previous year's interest, this process automatically compounds — no separate "compounding rule" needs to be memorised.

Figure

A staircase diagram showing the principal growing in three steps as each years interest is added before the next years interest is calculated

Derivation ACompound interest as repeated SI

The goal is to show that summing repeated simple-interest steps for a constant rate, year after year, produces exactly the compact formula $A = P\left(1+\dfrac{r}{100}\right)^n$ — so the formula is not a separate rule to memorise but the shortcut version of the year-by-year process already described above. We do this for a general $n$ by tracking the growth factor applied at each step.

Step 1. After Year 1, the amount is $A_1 = P_1 + SI_1 = P + \dfrac{P \times r \times 1}{100} = P\left(1+\dfrac{r}{100}\right)$ . Every principal, whatever its size, is multiplied by the same factor $\left(1+\dfrac{r}{100}\right)$ to reach the next year's amount — this factor is the heart of compounding.

Step 2. For Year 2, the new principal is $P_2 = A_1 = P\left(1+\dfrac{r}{100}\right)$ , and by the identical reasoning as Step 1, $A_2 = P_2\left(1+\dfrac{r}{100}\right) = P\left(1+\dfrac{r}{100}\right)^2$ . The growth factor has now been applied twice, once per year, and it multiplies rather than adds — this is what makes CI grow faster than SI over time.

Step 3. Repeating this once per year, each year multiplies the running amount by the same factor $\left(1+\dfrac{r}{100}\right)$ , so after $n$ years, $A_n = P\left(1+\dfrac{r}{100}\right)^n$ . This is exactly the standard compound interest formula, and the compound interest itself is $CI = A_n - P$ .

Step 4. This derivation assumes the rate $r$ stays the same every year — a constant factor is what allows the repeated multiplication to collapse into a single power. If the rate changes each year (say $r_1, r_2, r_3$ ), each year contributes a different factor, giving $A = P\left(1+\dfrac{r_1}{100}\right)\left(1+\dfrac{r_2}{100}\right)\left(1+\dfrac{r_3}{100}\right)$ , which no longer simplifies to a clean power — this is exactly why changing-rate problems must be solved by the repeated-SI method directly, one year at a time.

Worked example

Find the compound interest on ₹8,000 for 2 years at 5% per annum, compounded annually, by the step-by-step (repeated SI) method.

Step 1. Year 1: $P_1 = ₹8000$ . SI_1 = \dfrac{8000 \times 5 \times 1}{100} = ₹400 Amount at end of Year 1: $A_1 = 8000 + 400 = ₹8400$ .

Step 2. Year 2: new principal $P_2 = ₹8400$ . SI_2 = \dfrac{8400 \times 5 \times 1}{100} = ₹420 Amount at end of Year 2: $A_2 = 8400 + 420 = ₹8820$ .

Step 3. Total compound interest $= SI_1 + SI_2 = 400 + 420 = ₹820$ , so the compound interest is ₹820 and the final amount is ₹8,820. Notice Year 2's interest (₹420) exceeds Year 1's interest (₹400) even though the rate and time are identical — purely because the principal grew, exactly as the derivation above predicts.

Competitive stretch

A man borrows ₹12,000 for 3 years at 4% in Year 1, 5% in Year 2 and 6% in Year 3 (rate changing each year). Step 1. $P_1 = ₹12000$ , $SI_1 = \dfrac{12000\times4\times1}{100} = ₹480$ , $A_1 = ₹12480$ . Step 2. $P_2 = ₹12480$ , $SI_2 = \dfrac{12480\times5\times1}{100} = ₹624$ , $A_2 = ₹13104$ . Step 3. $P_3 = ₹13104$ , $SI_3 = \dfrac{13104\times6\times1}{100} = ₹786.24$ , $A_3 = 13104 + 786.24 = ₹13890.24$ . The amount to be repaid is ₹13,890.24, with total CI $= ₹1890.24$ — a sum only the repeated-SI method can handle, since the shortcut power formula needs one fixed rate throughout.

Common traps

Using the original principal every year, as in SI. In CI, Year 2's base is the amount at the end of Year 1, not the original $P$ .
Adding all the interest at the end instead of folding each year's interest into the principal before computing the next year's interest.
Expecting every year's interest to be equal. That is true only for SI; under CI, interest strictly increases each year whenever the rate is positive.
Rounding mid-calculation. When a rate produces a non-round number (like ₹786.24), carry the decimal through to the final step rather than rounding after each year.

Ask the tutor: Want to see why the repeated-SI method and the compound interest formula always give the identical answer for a constant rate, or try another changing-rate problem step by step? Ask the AI tutor.

Why can we not just use A = P(1+r/100)^n when the rate changes each year?

That formula comes from multiplying the same growth factor $(1+r/100)$ once per year, which only works if $r$ stays fixed. When the rate is $r_1$ in Year 1, $r_2$ in Year 2 and so on, each year contributes a different factor, so the correct amount is the product $P(1+r_1/100)(1+r_2/100)(1+r_3/100)\ldots$ , not a single power. The repeated-SI method builds this product automatically, one year at a time, so it always works regardless of whether the rate changes.

Is compound interest just simple interest applied three times in a row?

Almost, with one crucial twist: in SI applied three times, you would use the same principal $P$ each time and simply add the three interests. In CI, each application of the SI formula uses a different, larger principal — the amount left over from the previous year. It is "SI, repeated, but on a moving target," which is exactly why the total ends up bigger than three flat rounds of ordinary SI on the same $P$ .

Do I have to use the repeated-SI method even when the rate is constant, or can I jump straight to the formula?

Once you trust the formula, you can jump straight to $A = P(1+r/100)^n$ for constant-rate problems — it is faster. But ICSE examiners frequently ask you to show the year-by-year working explicitly, especially for 2- and 3-year sums, to check you understand why the formula holds and not just that you can plug numbers into it. Practising the repeated-SI method also makes changing-rate problems, where the formula cannot be used, much less intimidating.

Compound interest as repeated SI

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads