Question

If 3000 dollars is invested in a bank account at an interest rate of 4 per cent per year,

Find the amount in the bank after 6 years if interest is compounded annually:
Find the amount in the bank after 6 years if interest is compounded quarterly:
Find the amount in the bank after 6 years if interest is compounded monthly:
Finally, find the amount in the bank after 6 years if interest is compounded continuously:

Answer 1

`~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+(r)/(n)\right)^(nt) \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$3000\\ r=rate\to 4\%\to (4)/(100)\dotfill &0.04\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{\underline{annually}, thus once} \end{array}\dotfill &1\\ t=years\dotfill &6 \end{cases} \\\\\\ A = 3000\left(1+(0.04)/(1)\right)^(1\cdot 6) \implies A \approx 3795.96 \\\\[-0.35em] ~\dotfill`

`~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+(r)/(n)\right)^(nt) \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$3000\\ r=rate\to 4\%\to (4)/(100)\dotfill &0.04\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{\underline{quarterly}, thus four} \end{array}\dotfill &4\\ t=years\dotfill &6 \end{cases} \\\\\\ A = 3000\left(1+(0.04)/(4)\right)^(4\cdot 6) \implies A \approx 3809.20 \\\\[-0.35em] ~\dotfill`
`~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+(r)/(n)\right)^(nt) \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$3000\\ r=rate\to 4\%\to (4)/(100)\dotfill &0.04\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{\underline{monthly}, thus twelve} \end{array}\dotfill &12\\ t=years\dotfill &6 \end{cases} \\\\\\ A = 3000\left(1+(0.04)/(12)\right)^(12\cdot 6) \implies A \approx 3812.23 \\\\[-0.35em] ~\dotfill`

`~~~~~~ \textit{Continuously Compounding Interest Earned Amount} \\\\ A=Pe^(rt)\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill & \$3000\\ r=rate\to 4\%\to (4)/(100)\dotfill &0.04\\ t=years\dotfill &6 \end{cases} \\\\\\ A = 3000e^(0.04\cdot 6) \implies A \approx 3813.75`