solution to series I'm trying to solve these two equations:
$$12\times{}y = \sum_{m=0}^{11}\sum_{n=0}^{m}r\times(1+r-r\times{}t)^n$$
$$12\times{}y\times{}s = \sum_{m=0}^{11}\sum_{n=0}^{m}r\times{}t\times{}(1+r-r\times{}t)^n$$
$y$ and $s$ are known. $r$ and $t$ are unknown. I do not know how to proceed. I cannot see a way to simplify or isolate the unknowns. Does anyone have an idea how to proceed?
 A: First, note that if you divide the LHS of the second equation by the LHS of the first, you just get $s$. If you divide the RHS of the second equation by the RHS of the first, you just get $t$. Since the two sides are equal in both equations, the ratios are both of the same two numbers, and therefore are the same. That is, $t = s$. (Note that this doesn't work when $y=0$, because then you would be dividing by $0$. When $y = 0$, $t$ can have any value.)
To solve for $r$ we use GNU Supporter's comment. The sum of a finite geometric series is 
$$\sum_{i=0}^k a^i = \frac{a^{k+1} - 1}{a - 1}$$
But to simplify the notation a bit, let $u = 1 - t = 1- s$, then the first equation becomes
$$12y = \sum_{m=0}^{11}\sum_{n=0}^mr(1 + ru)^n$$
Now $$\sum_{n=0}^mr(1 + ru)^n = r\frac{(1 + ru)^{m+1} - 1}{ru} = \frac 1u((1 + r - tr)^{m+1} - 1)$$
and $$\begin{align}\sum_{m=0}^{11}\sum_{n=0}^mr(1 + ru)^n &= \frac 1u\sum_{m=0}^{11}\left((1 + ru)^{m+1}-1\right)\\&=\frac 1u\left(\sum_{m=1}^{12}(1 + ru)^{m}-12\right)\\&=\frac 1u\left(\sum_{m=0}^{12}(1 + ru)^{m}-13\right)\\&=\frac1u\left(\frac{(1 + ru)^{13} - 1}{ru}-13\right)\\&=\frac{(1 + ru)^{13} - 1}{ru^2}-\frac{13}{u}\end{align}$$
Therefore $$12y = \frac{(1 + ru)^{13} - 1}{ru^2}-\frac{13}{u}\\12yru^2 = (1 + ru)^{13} -13ru$$
which is a 13-degree polynomial in $r$. You can simplify it a bit by making the substitution $v = 1 + ru$ and noting that $ru = 1-v$, so
$$12yu(1-v) = v^{13} - 13(1-v)\\v^{13} + (13+12yu)v -(13+12yu) = 0$$
But that is the best you can do exactly. There is no nice "quadratic formula"-style way of expressing the roots of a 13-degree polynomial, so you would have to solve this by approximation techniques such as Newton's method. I can tell you that it has 1 to 3 real roots. The remaining roots will all be complex numbers.
Once you do obtain a root for $v$, you can recover $$r = \frac{1 - v}u = \frac{1-v}{1-s}$$
