Question

An initial investment of $1000 is appreciated for 8 years in an account that earns 9% interest, compounded annually. Find the amount of money in the account at the end of the period

Answer 1

B{0} = 1000
\begin{align}
B_{n}& = B_{n-1} + (.09)B_{n-1} \\
&=B_{n-1}(1+.09)\\
&=B_{n-1}(1.09)
\end{align}

\begin{align}
B_{n}&=B_{n-1}(1.09)\\
&=B_{n-2}(1.09)(1.09)\\
&=B_{n-3}(1.09)(1.09)(1.09); or \ simply:\ B_{n-3}(1.09)^3\\
&=B_{n-4}(1.09)^4\\
&=B_{n-r}(1.09)^r\\
\end{align}
When n-r = 0, we get back to B{0}; n-r = 0 when r=n
\begin{align}
B_{n}&=B_{n-r}(1.09)^r; r=n\\
&=B_{n-n}(1.09)^n\\
&=B_{0}(1.09)^n;\ B_0=1000
\end{align}
\begin{align}
B_n&=1000(1.09)^n;\ where \ n=number\ of\ cycles
\end{align}

Answer 2

8 years times one per year equals 8 cycles

B{8} = 1000(1.09)^8