Answer:
Simple interest
![[P+P* (r)/(100) * n]](https://img.qammunity.org/2020/formulas/mathematics/middle-school/sy6e07zrsi3ynwg22x310alxkqjdif8chj.png)
Compound interest
![P[1+(r)/(100) ]^(n)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/jwyztfj8eflmzdm4u9zgp139ix3n8nlue3.png)
Explanation:
If a certain sum of money P is increasing at a rate of r% simple interest annually, then after n years the increases sum will become
![P[1+(r)/(100)* n]\textrm {i.e.}[P+P* (r)/(100) * n]](https://img.qammunity.org/2020/formulas/mathematics/middle-school/z14v0k98mqsh93hmu2805f7it6wcgwuuuy.png)
Therefore, in simple interest the interest is calculated on the fixed principal amount P.
So, after 1 year the sum will become
![[P + P * (r)/(100) ]](https://img.qammunity.org/2020/formulas/mathematics/middle-school/bupfrk86meadkrt5j3af2wizov5aj1ftbr.png)
After 2 years the sum will become
![[P + (P * (r)/(100))+(P * (r)/(100)) ]](https://img.qammunity.org/2020/formulas/mathematics/middle-school/pnaauo4fw6b803deceo0v4tsb4rb4s8nf6.png)
Therefore, in each year the sum is increasing by a fixed amount and it is the simple interest which is calculated on the principal P always.
But, if we consider P is increasing at a rate of r% interest annually and is compounded every year then after n consecutive years the sum will become
.
So, after first year the sum will become
.
After 2 years the sum will become
![[P + P * (r)/(100) ] +[P + P * (r)/(100) ] * (r)/(100)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/cq8dqlbu8r9iwm213r5eubc9ectf6lp7kh.png)
Therefore, in the 2nd year the interest is calculated on
i.e. the principal after one year but not on P only.
Similarly the interest after 3rd year will be calculated on the principal that becomes after 2 years. (Answer)