How do I calculate a multiplier over X iterations?

Question

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 = 1
Sum Of All Numbers = 1000
Number of increases = 99
Multiplier = X

Any help would be very appreciated.

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EDIT: I forgot an important detail. DUH! Sorry.

I want the sum of all of the 100 numbers to equal 1000 or less.

Answer 1

$$x^{100} = 1000$$ $$x = 1000^{1/100} \approx 1.0715$$

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Response to comment: $$1+x + x^2 + \cdots + x^{100} = 1000$$ $$\frac{x^{101} - 1}{x-1} = 1000$$ WolframAlpha suggests the solution is near $1.0365$

Answer 2

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.

Answer 3

$$ \begin{array}{ccccccccccccccl} 1 & + & X & + & X^2 & + & X^3 & + & \cdots & + & X^{99} & & & = & 1000 \\ & & X & + & X^2 & + & X^3 & + & \cdots & + & X^{99} & + & X^{100} & = & 1000 X \end{array} $$ 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.