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Determine how much you will have to save each month at $3$, $6$, $9$, and $12$ percent compounded monthly for you to accumulate a nest egg for retirement. The variables are current age, age of retirement, nest egg size, and interest rate. Show all work and cite as appropriate.

C = current age A = Age of retirement R = interest rate N = nest egg size P = Cash flow per period T = Time $N=P(1+\frac{.03}{12})^{12}(55-24)$

$\$6583.33$ per month at $3$ percent

can someone please tell me if i did this right and my answer is accurate or not?

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    Hi, welcome to math.stackexchange! Please note that the tags appear wherever the title appears (e.g. on the main page), so repeating the overall field of the question (precalculus) is redundant. The title should summarize the question more specifically, e.g. "Compound interest for retirement".2012-08-09
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    I don't understand the calculation (mostly because you introduce lots of variables and then never use them, and because the formatting is confusing), but from what I know about retirement I doubt that \$6583.33 per month is the right answer :-) Note that you can enclose formulas in dollar signs to get $\TeX$ formatting (in this case the raised exponent) -- single dollar signs for inline formulas and double dollar signs for displayed equations. P.S.: I noticed that the variables are on individual lines in the source. Note that single line breaks are ignored; you need double breaks for newlines.2012-08-09
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    Not right, as a commonsense look at the numbers will show, it is far too big. You need to find the **sum** of a geometric series. For very fine detail, should specify whether deposits now and then monthly up to a month *before* retirement, or something a bit different. Recall that the sum of the geometric series $1+x+\cdots+x^{n-1}$ is $(x^n-1)/(x-1)$. You seem to have only calculated what a deposit **now** grows to. What about all the other payments?2012-08-09
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    do u maybe happen to know how i could possible do it?2012-08-09
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    I indicated it in the comment above. Let $x=1+\frac{0.3}{12}$. The payment $P$ a month before retirement grows to $Px$. The one two months before grows to $Px^2$, The one before that to $Px^3$. The first one to $Px^{372}$. Find sum $x+x^2+\cdots+x^{372}=Q$. If $N$ is nest egg, monthly payment is $N/Q$. If pay first a month from now, and make a payment on retirement day, $Q=1+x+\cdots +x^{371}$.2012-08-09
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    this is a different question but does 200000=P(1+.03/12)^12(55-24) = 6583.33? im not sure if i am doing this right2012-08-09
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    I **think** you are asking what $P$ is if $200000=P\left(1+\frac{0.03}{12}\right)^{12(55-24)}$. Nowhere near $6583.33$. The answer is more like $79,002$. If you divide this by $12$ you get roughly $6583.3$. That is probably what you did.2012-08-09

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One needs to specify the payment scheme a bit more accurately, because for exact computation we need those details. Let $n$ be the number of months between current age and retirement age. I will assume that $n$ is an integer.

I will also assume that you make a payment $P$ every month, with first payment today, and the last payment done a month before retirement, for a total of $n$ payments. You will have to make minor adjustments if the payment scheme is slightly different. Let $N$ be the desired sum available at retirement.

Let $r$ be the nominal yearly rate. I assume we have monthly compounding. The monthly rate is $\frac{r}{12}$. Let $x=1+\frac{r}{12}$.

So in one month $1$ unit of currency grows to $x$ dollars. The payment $P$ you made a month before retirement has grown to $Px$ at retirement, not much! The one you made two months before retirement has grown to $Px^2$. The one three months before retirement has grown to $Px^3$. And so on. Finally, the one you made $n$ months before retirement has grown to $Px^n$. (The payments made long before retirement have grown quite a bit.)

We want to have accumulated a total of $N$, so $$N=Px+Px^2+Px^3+\cdots +Px^n=Px(1+x+x^2+\cdots+x^{n-1}).$$ The sum of the geometric series $1+x+x^2+\cdots+x^{n-1}$ is $\frac{x^n-1}{x-1}$. (This is a standard formula, you can look it up on Wikipedia.) So our equation becomes $$N=Px\frac{x^n-1}{x-1}.$$ Now it is all ready for numerical calculation. With interest rate $3\%$, for example, $x=1.0025$.