During the first year the account will earn 5% of $8000, or \(0.05 (8000) = 400\) dollars. Thus, at the end of the first year the account will contain the original $8000 plus the $400 interest for a total of $8400.
At the end of the second year the account will have the $8400 from the previous year, plus 5% of $8400, or \(0.05(8400) = 420\) dollars in interest. We can write this sum as
\begin{equation*}
8400 + 8400(0.05) = 8400 (1 + 0.05) = 8400(1.05)~~ \text{dollars.}
\end{equation*}
Thus, at the end of the second year there will be \($8400(1.05) = $8820\) in the account.
In fact, each new balance is found by multiplying the previous balance by 1.05. The balances at the ends of the third and fourth years are
\begin{equation*}
\$8820(1.05) = \$9261~~~~\text{and}~~~~\$9261(1.05) = \$9724.05
\end{equation*}
The annual balances form the sequence 8400, 8820, 9261, and 9724.05, as shown below.
\(n\) |
\(a_n\) |
1 |
8400 |
2 |
8820 |
3 |
9261 |
4 |
9724.05 |