Krdytkyu

Let's say I want to distribute leaflets for 60 months and in total I want to distribute $2500000$ leaflets. In my first month I want to distribute $8000$ leaflets and then I want to increase my speed of distribution every month at the same rate to finally reach $2500000$ at month 60.

Essentially I think I want this equation (1):
$$8000+8000x+8000x^2+...8000x^{59}=2.50cdot10^6$$
We can rewrite it as
$$8000[1+x+x^2+...+x^{59}]=2.5cdot10^6$$
Hence
$$1+x+x^2+...x^{59}=312.5$$
I have a geometric series with $a_1$=1 and I need to find r. Hence
$$312.5=frac{1cdot(1-x^{60})}{1-x}$$
Which gives the polynomial (2):
$$x^{60}-312.5x+311.5=0$$
I plug into matlab and get a bunch of complex roots and two real roots which are:

1.046923161434767 + 0.000000000000000i

1.000000000000000 + 0.000000000000000i

I double check that the first root works in excel (Buncf or rows hidden) and it works HAPPY DAYS:

QUESTION
I undersrtand that $x=1$ is a root of the polynomial (2), but clearly it does not satisfy my initial equation (1). Why did the equation for the sum of finite geometric series allow this thing? If I were to use any of the remaining 58 complex roots, would it give me 2.5 million?

You multiplied by $1-x$ when you converted the fractional equation to a polynomial equation. Then a root $1-x=0, x=1$ comes from this multiplying factor and not your original equation.

Had you stayed with the original geometric series sum, which had degree $59$ instead of $60$, there would not have been this extra multiplying factor and all of the roots would be good, except for the fact that distributing a complex number of leaflets might be a little challenging to implement. Then again, entering the long polynomial into the computer is less challenging but still not attractive.

The formula for the sum of a finite geometric series is not valid when the common ratio $x$ is $1$. Indeed, it is derived by dividing by $1-x$, which is not possible if $x=1$ So, the equation $$312.5=frac{1cdot(1-x^{60})}{1-x}$$ is only equivalent to your original equation (1) assuming that $xneq 1$. Under that assumption, you can then reversibly multiply both sides by $1-x$ to find that equation (2) is equivalent.

So, equations (1) and (2) are equivalent assuming $xneq 1$. This means that the solutions to equation (2) besides $xneq 1$ are all solutions to equation (1), and $x=1$ may or may not also be a solution to equation (1) (and in this case it is not).

More generally, the lesson here is that solving an equation is not just a matter of doing manipulations to reach an answer. It is crucial to understand the logic behind these manipulations, since you are really making an argument that certain equations imply other equations under certain assumptions. The directionality of the implications and the assumptions needed make all the difference.