What is the general formula, if any, for the following sum of increasing ratios?
$$
\frac{i}{c}+\frac{a}{c(1-x)}+\frac{ab}{c(1-2x)}+\frac{ab^2}{c(1-3x)}+\cdots+\frac{ab^n}{c(1-(n+1)x)}
$$
Each subsequent numerator with $a$ is the previous one multiplied by a factor $b$; ie, for a factor of 1.5 the sequence would be $50, 75, 112.5,\ldots$
$i$ is just an arbitrary numerator for the first fraction. All the numerators can be added together involving a geometric sum:
$$i+a \Bigl(\frac{b^{n+1}-1}{b-1}\Bigr)$$
Each subsequent denominator is $c$ reduced in steps of $x$ percent in decimal form. For example, if $c$ is 555 in the first fraction and $x$ is 0.04 the following denominators would be 532.8, 510.6, 488.4… All denominators summed up contain the sum of all natural numbers up to $n+1$:
$$c\Bigl[n+2-x\Bigl(\frac{n^2+3n+2}{2}\Bigr)\Bigr]$$
But we know that
$\frac{m+p}{n+q} \ne \frac{m}{n}+\frac{p}{q}$. Yet, since there are formulas for the sum of the numerators and the denominators, -as shown- there might be one to express their ratio.
Best Answer
As said in comments, you made quite strange (and wrong) manipulations.
Hoping that you enjoy special functions $$\sum_{k=0}^n \frac{a b^k}{c (1-(k+1) x)}=\frac a c\sum_{k=0}^n \frac{ b^k}{ 1-(k+1) x}$$ $$\sum_{k=0}^n \frac{ b^k}{ 1-(k+1) x}=\frac1 x \Bigg[b^{n+1} \Phi \left(b,1,n+2-\frac{1}{x}\right)-\Phi \left(b,1,1-\frac{1}{x}\right) \Bigg]$$ where appear the Lerch transcendent function.