Suppose we have a population of bacterial that grows by 10 % every day. Suppose right now we have 10 000 bacteria count then in 10 days how many bacteria we got??
Attempt
Let $A_0 = 10000$ be initial amount. So, $A(1) = A_0 + A_0\frac{1}{10} $ and $A(2) = A_0 + \frac{1}{10} A_0 + \frac{1}{10} A_0 + \frac{1}{100} A_0 $. and so on... but after $10$ years the formula will get messy. Is it a simpler way to do this?
Call the population at the end of the $n$th day $P(n)$.
A $10 \%$ increase means multiplying the previous population by $1.1$
At the start, $P(0) = 10000$
At the end of day $1$, $P(1) = 1.1(10000)$
At the end of day $2$, $P(2) = (1.1)(1.1)(10000) = 1.1^2(10000)$
...
At the end of day $k$, $P(k) = 1.1^k(10000)$
Can you proceed?