How do I calculate a multiplier over X iterations? Let's say I have the number 1, and I want to increase it 99 times until the sum of all 100 numbers (including the original 1) equals 1000. I want the multiplier to remain constant, and to be used on the result of the previous number.
So...
1 * 1.1    = 1.1
1.1 * 1.1  = 1.21
1.21 * 1.1 = 1.331
and so on...
Basically, the multiplier is X since all of the other parts are known, but I am totally stuck.
Starting Number = 1Sum Of All Numbers = 1000Number of increases = 99Multiplier = X
Any help would be very appreciated.

EDIT:
I forgot an important detail. DUH! Sorry.
I want the sum of all of the 100 numbers to equal 1000 or less.
 A: $$
\begin{array}{ccccccccccccccl}
1 & + & X & + & X^2 & + & X^3 & + & \cdots & + & X^{99} & & & = & 1000 \\
& & X & + & X^2 & + & X^3 & + & \cdots & + & X^{99} & + & X^{100} & = & 1000 X
\end{array}
$$
Subtract the first row above from the second, getting
$$
X^{100} - 1 = 1000X - 1000
$$
$$
X^{100}-1 = 1000(X-1)
$$
$$
1000 = \frac{X^{100}-1}{X-1}
$$
Possibly solving this for $X$ can be done only numerically.
$X=1.0370627$ seems to come pretty close.
A: $$x^{100} = 1000$$
$$x = 1000^{1/100} \approx 1.0715$$

Response to comment:
$$1+x + x^2 + \cdots + x^{100} = 1000$$
$$\frac{x^{101} - 1}{x-1} = 1000$$
WolframAlpha suggests the solution is near $1.0365$
A: You have a geometric progression.  If the first term is $a$ and the common ratio is $r$, the $n^{th}$ term is $ar^{n-1}$ because you have multiplied by $r\ n-1$ times.  In your case you want $a=1, ar^{100}=1000$, so
$$r^{100}=1000\\100 \log(r)=\log(1000)\\ \log(r)=\frac {\log(1000)}{100}\\r=e^{\frac {\log(1000)}{100}}\approx 1.071519$$
Note that when you ask for $100$ increases you have $101$ terms.
