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Suppose you want to accumulate $12,000$ in a $5 \%$ account by making a level deposit at the beginning of each of the next 9 years. Find the required level payment. So this seems to be an annuity due problem. I know the following:

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

So in this problem, we are trying to solve for $P$? Just plug in the numbers? Or do we need to calculate the discount rate $d = i/(i+1)$ since the annuity is being payed at the beginning of the year?

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Does 'level deposit' mean that you are depositing the same amount into this account each year? –  WWright Oct 16 '10 at 19:50
    
$P$ is the Principal, Present Value or Present Worth and $A$ is the Annuity. In my opinion you have to find a formula for the Future Value $F=12000$ as a function of $A$, the interest rate $i$ and the number of periods $n$. –  Américo Tavares Oct 16 '10 at 19:59
    
Well $P = \frac{F}{(1+i)^{n}}$. Is that it? So $F= P(1+i)^n$ and then just solve for $P$ using the above geometric series? So we get $F = A \frac{(1+i)^n-1}{i(1+i)^n}(1+i)^n$ –  PEV Oct 16 '10 at 23:58
    
This simplifies to $F = A \frac{(1+i)^{2n}-(1+i)^n}{i(1+i)^n}$. Is that correct? –  PEV Oct 17 '10 at 0:05
    
I repeat here my comment below. The accumulated amount $F_{n}$ (at the beginning of year $n$) of $n$ Annuities $A$ capitalized at the interest rate of $i$ is given by $F_{n}=\displaystyle\sum_{k=1}^{n}A(1+i)^{n-k}=\displaystyle\sum_{m=1}^{n}A(1+i)‌​^{m-1}=A\dfrac{(1+i)^{n}-1}{i}$. Hence $A=F_{n}\dfrac{i}{(1+i)^{n}-1}$. At the end of year $n$, the total accumulated amount is $F_{n+1}=\left( 1+i\right) F_{n}$. Thus $A=F_{n+1}\dfrac{i}{(1+i)^{n+1}-\left( 1+i\right) }$. For $n=9$, $A=\dfrac{12000\times 0.05}{1.05^{10}-1.05}\approx 1036.5$ –  Américo Tavares Oct 17 '10 at 14:10
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2 Answers

The problem statement is missing the time when you want to have the 12,000. If it is at the end of the ninth year, the value of the deposit at the beginning of year n will have increased by 1.05^(10-n). So if A is the deposit you have $\displaystyle \sum_{k=1}^{n} A*1.05^{(10-k)}=12000$. Solve for A

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I get $F = A \frac{(1+i)^{2n}-(1+i)^n}{i(1+i)^n}$. But when I solve for $A$, I get the level payment as being $1088.28$. But the correct answer is $1036.46$. Why is there a discrepancy? –  PEV Oct 17 '10 at 0:14
    
+1 for the exposition. –  Américo Tavares Oct 19 '10 at 16:19
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I get $F = A \frac{(1+i)^{2n}-(1+i)^n}{i(1+i)^n}$. But when I solve for $A$, I get the level payment as being $1088.28$. But the correct answer is $1036.46$. Why is there a discrepancy?

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The difference is 5%. This represents the interest in the last year. I think your answer assumes you withdraw the whole amount just after the ninth deposit and the problem assumes it stays for one more year. –  Ross Millikan Oct 17 '10 at 0:44
    
As Ross wrote, "the difference is 5%". Here is my explanation. The accumulated amount $F_{n}$ (at the beginning of year $n$) of $n$ Annuities $A$ capitalized at the interest rate of $i$ is given by $F_{n}=\displaystyle\sum_{k=1}^{n}A(1+i)^{n-k}=\displaystyle\sum_{m=1}^{n}A(1+i)‌​^{m-1}=A\dfrac{(1+i)^{n}-1}{i}$. Hence $A=F_{n}\dfrac{i}{(1+i)^{n}-1}$. At the end of year $n$, the total accumulated amount is $F_{n+1}=\left( 1+i\right) F_{n}$. Thus $A=F_{n+1}\dfrac{i}{(1+i)^{n+1}-\left( 1+i\right) }$. For $n=9$, $A=\dfrac{12000\times 0.05}{1.05^{10}-1.05}\approx 1036.5$ –  Américo Tavares Oct 17 '10 at 14:06
    
This is the same thing as an annuity due right? In which case the present value is: $\frac{1-v^n}{d}$ and the future value is $(1+i)^{n} \frac{1-v^n}{d}$ where $v = 1/(1+i)$ and $d = i/(i+1)$. –  PEV Oct 17 '10 at 14:09
    
Your formulae are the same, apart from the missing $A$. I would say that: (i) we pay 9 annuities at the beginning of each year, and (ii) the Future value (accumulated amount) is received at the end of year 9. –  Américo Tavares Oct 17 '10 at 14:33
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