Compound Interest: Formulas

Compound interest differs from Simple interest as the interests earned in the previous years considered as principal for the next years.  Observe the table From the above table, we can see that in Simple interest case, Interests earned in each year is same, but in the compound interest, except for the first year, interests are different.  This is due to the fact that in the CI case interest earned in the 1st year is considered as principal and interest will be calculated on this too.  In the first box 10+1 can be explained as 10 is the interest on the principal, 1 is the interest on the first year interest.  For the 2nd box, 10 is the interest on principal, 1 is interest on first year interest, another 1 is interest on second year 10, and 0.1 is the interest on 1.

Compound Interest formula:

There is no direct formula for Compound interest but $P{\left( {1 + \displaystyle\frac{R}{{100}}} \right)^n}$ gives us the Amount which includes principal as well as interest.  So to find the interest we need to substract principal from the amount.
CI = A - P = $$P{\left( {1 + \dfrac{R}{{100}}} \right)^n} - P$$

Compound interest through Pascal Triangle:

Find the compound interest on Rs.6000 at the rate 10% per annum for 3 years.
Formula: A = $P{\left( {1 + \displaystyle\frac{R}{{100}}} \right)^n}$ = $6000{\left( {1 + \displaystyle\frac{{10}}{{100}}} \right)^3}$
$\Rightarrow 6000 \times {\left( {\displaystyle\frac{{11}}{{10}}} \right)^3}$ = 7986
Compound Interest = 7986 - 6000 = 1986

Alternate method:
We can solve compound interest problems easily with the use of pascal triangle. To solve above problem, we need to identify the coefficients where 3 appears first time in a row.  We know that in 4th row 3 appears.  We need to take coefficients excluding 1.  i.e., 3, 3, 1. Now CI = 3 × (10% of 6000) + 3 × (10% of 600) + 1 × (10% of 60) = 1986

Finding compound interest when periods ($$n$$) is not an integer:

For example, we have to find compound interest on Rs.2000 at 15% interest for 2 years 4 months.
There is no difference in calculation for the 2 years. But for 4 months, take 4 months interest.
4 months =$$\dfrac{4}{{12}} = \dfrac{1}{3}$$rd of an year. So interest rate for the last 4 months = $$\dfrac{1}{3}\left( {15} \right)$$
In this case, A = $$2000{\left( {1 + \dfrac{{15}}{{100}}} \right)^2}\left( {1 + \dfrac{{{\textstyle{1 \over 3}}\left( {15} \right)}}{{100}}} \right)$$
= $$2000{\left( {1 + \dfrac{3}{{20}}} \right)^2}\left( {1 + \dfrac{5}{{100}}} \right)$$
= $$2000{\left( {\dfrac{{23}}{{20}}} \right)^2}\left( {1 + \dfrac{1}{{20}}} \right)$$
= $$2000{\left( {\dfrac{{23}}{{20}}} \right)^2}\left( {\dfrac{{21}}{{20}}} \right)$$
= $$\dfrac{{11109}}{4}$$ = $$2777.25$$
So compound interest = 2777.25 - 2000 = 777.25

To find the difference between compound interest and simple interest for two years :

Compound interest for two years = $$p{\left( {1 + \dfrac{R}{{100}}} \right)^2} - p$$
Simple interest for two years = $$p \times 2 \times \dfrac{R}{{100}}$$

=$$\left[ {p{{\left( {1 + \dfrac{R}{{100}}} \right)}^2} - p} \right] - \left( {p \times 2 \times \dfrac{R}{{100}}} \right)$$
= $$p\left[ {{{\left( {1 + \dfrac{R}{{100}}} \right)}^2} - 1 - \left( {\dfrac{{2R}}{{100}}} \right)} \right]$$
= $$p\left[ {1 + \dfrac{{2R}}{{100}} + {{\left( {\dfrac{R}{{100}}} \right)}^2} - 1 - \dfrac{{2R}}{{100}}} \right]$$
= $$p{\left( {\dfrac{R}{{100}}} \right)^2}$$
Hence, difference between compound interest and simple interest @R % p.a. for two years = $$p{\left( {\dfrac{R}{{100}}} \right)^2}$$

To find the difference between compound interest and simple interest for three years :

Use the following formula.
Difference = $$p{\left( {\dfrac{R}{{100}}} \right)^2}\left( {3 + \dfrac{R}{{100}}} \right)$$
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