An investment of 1,000,000 receives a 10% bonus every year. A total of $48,000 is withdrawn from the investment each year. Assume that you have got the 10% bonus before you withdraw each year.
I just got the recurrence relation which is s(0)=1000000,and s(n)=1.1xS(n-1)-48000. I need the iterative approach (or so call the explicit formula ). Thanks.
Let $T=1,000,000$, $W=48 ,000$, and $q=1.1$. With these notations we have: $$ \begin{matrix} \text{ after the first withdrawal:}& T q -W\\ \text{ after the second withdrawal:}& Tq^2 -Wq-W\\ \text{ after the third withdrawal:}& Tq^3 -Wq^2-Wq-W\\ \vdots\\ \text{ after the }\ n^{th}\ \text{withdrawal:}& Tq^n -W\sum_{i=0}^{n-1}q^i. \end{matrix} $$
A closed form of the sum above can be obtained like this. (Let $S_n=\sum_{i=0}^{n-1}q^i$.) Then
$$S_n-qS_n=1-q^n,$$
or
$$S_n=\frac{\ 1-q^n}{1-q}.$$ So, after the $n^{th}$ withdrawal, we have
$$Tq^n-W\frac{\ 1-q^n}{1-q}.$$