What is the derivation for the formula that gives the difference between compound interest and simple interest after three years: $P\left(\frac R{100}\right)^2 \left(\frac R{100} + 3\right)$? It is the formula for C.I - S.I for 3 years which I read in many different places but I'm not able to figure out how it is derived.
Can anyone help?
On
The formula for simple interest is
\begin{align*} A = P \bigg(1 +\dfrac{n R}{100}\bigg)\tag{1} \quad\text{where}&\\ &\text{A = final amount after interest}\\ &\text{P = original principal}\\ &\text{n = number of cycles for interest rate}\\ &\text{R = annual interest rate}\\ \text{but compound interest is}&\\ &A=P\bigg(1+\frac{R}{100}\bigg)^{\Large{n}}\\ \tag{2} \text{ compounding once/year}&\\ \text{or}&\\ &A = \space P\bigg(1+\dfrac{R}{100n}\bigg)^{\Large{nt}}\\ &\text{n is the number of payments per year}\\ &\text{t = number of years} \end{align*}
I believe what you want is the difference between either of $\quad(2)\space$ minus $\space(1)$
The first thing we do, let's kill all the percentages. Calculations like this are much easier when interest rates are given as pure numbers, for example $0.02$ instead of $2$%.
So if you start with principal $P$ and leave it invested at a rate $r$ (where $r = \frac R{100}$) for three years at simple interest, you end up with $P(1 + 3r)$ at the end.
But if you earn compound interest, the amount at the end is $P(1 + r)^3$.
Now take the difference in the outcomes:
$$\begin{eqnarray} P(1 + r)^3 - P(1 + 3r) &=& P(1 + 3r + 3r^2 + r^3) - P(1 + 3r) \\ &=& P(3r^2 + r^3) \\ &=& P(3 + r)r^2. \end{eqnarray}$$
Now that we know the answer, we can put it back in terms of percentages if we must:
$$P(3 + r)r^2 = P \left(3 + \frac{R}{100}\right) \left(\frac{R}{100}\right)^2.$$