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To preface this, it's been years since my last calc class, but I feel like this is lower level than that. I am looking for the notation for an equation that sums up the values of the same equation iterated several times, while carrying the last result through the "new" iteration.

I am a programmer at heart, so using "for" statements seems to accomplish this pretty easy, but I need an equation that represents it as well.

So here's the idea: Let's say that we start with 500 people in a church, and over the course of 2 years we assume that each one of those people bring 2 more people who become members which gives you a new church population of 1500, then iterate it again until 6 years have passed.

By the end you should have 13,500 members. It's funny that it's so easy to do in your head, but for the life of me I have no idea how to represent this in an equation. I have tried using a summation to represent the equation using x=2x+x, but I can't figure out how to "notate" x equaling the value of the last iteration with each "new" iteration.

Does this make any sense? I am probably completely wrong trying to use summation notation to do this, but I don't what else to try.

Thanks for any advice.

joshrathke
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  • Let $x_n$ be the population after $2n$ years, Then $x_{n+1}=2x_n+x_n=3x_n$. (This is the mathematical version of your programmer's $x=2x+x$.) So if $x_0=500$, then $x_1=(3)(500)$, $x_2=(3)(3)(500)=(3^2)(500)$, and in general $x_n=(3^n)(500)$. – André Nicolas May 09 '15 at 04:05

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You have a geometric progression assuming each person only brings in two others and then stops. In that case you have a tree-you start with $n$, each brings in two so you add $2n$, each of those brings in two so you add $4n$, so you have $n(1+2+4+\dots 2^k)=n(2^{k+1}-1)$ for $k$ doublings. If the original members keep bringing in newbies, you have a different relation. It is harder if the originals bring in some members in year 1 who start bringing in newbies in year 2. Your question is not specific enough for that.

Ross Millikan
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