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Suppose you are given $\\$\\var{C}$ to deposit into a savings account at the end of each {period[2]} for $\\var{years}$ years. If the interest rate is $\\var{ipa}\\%$ per annum compounding {period[0]}, how much money is in the savings account after $\\var{years}$ years?

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You are asked to find the future value of an ordinary annuity (since the payments are at the end of each period). Therefore we will use the future value of an ordinary annuity formula

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$\\displaystyle A=R(\\frac{(1+i)^n-1}{i})$

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where $A$ is the future value, $R$ is the amount deposited per period, $i$ is the interest rate per period, and $n$ is the number of periods.

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In our situation we have,

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$R=\\var{C}$,

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$i=\\frac{\\var{ipa}\\%}{\\var{period[1]}}=\\frac{\\var{ipadec}}{\\var{period[1]}}$, $i=\\var{ipa}\\%=\\var{ipadec}$,

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$n=\\var{years}\\times \\var{period[1]}=\\var{n}$, $n=\\var{n}$,

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and therefore we have

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$\\displaystyle A=\\var{C}(\\frac{(1+\\frac{\\var{ipadec}}{\\var{period[1]}})^\\var{n}-1}{\\frac{\\var{ipadec}}{\\var{period[1]}}})$

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Calculating this we find 

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$\\begin{align}A&\\approx \\var{F}\\\\&=\\$\\var{Frounded}\\quad \\text{(to the nearest cent)}\\end{align}$

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Which formula should we use to calculate the future value of the annuity in $\\var{years}$ years?

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Amount in the savings account after $\\var{years}$ years:

\n

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$\\$$ [[0]] (to the nearest cent)

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You are asked to find the future value of an ordinary annuity (since the payments are at the end of each period). Therefore we will use the future value of an ordinary annuity formula

\n

$\\displaystyle A=R\\left[\\frac{(1+i)^n-1}{i}\\right]$

\n

where $A$ is the future value, $R$ is the amount deposited per period, $i$ is the interest rate per period, and $n$ is the number of periods.

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