Consecutive powers (Part 5)

First, we should know that the validity of this infinite expansion is only when $|x|<1$ . Let us check that out.

$x+x^2 + x^3 + \ldots = \dfrac{x}{1-x}$ $999=\dfrac{x}{1-x}$ $x=\dfrac{999}{1000}$

So, this is true, since it is less than $1$ .

So, the answer is $\boxed{\dfrac{999}{1000}}$ .

Here is the easy way to find $x$ by factoring:

$x+x^{2}+x^{3}+x^{4}+...=999$

$x[1+(x+x^{2}+x^{3}+x^{4}+...)]=999$

$x(1+999)=999$

$1000x=999 \rightarrow x= \boxed{\frac{999}{1000}}$

If $|x|<1$

$a_{1}+a_{1}x+a_{1}x^2+a_{1}x^3+\cdots =\frac{a_{1}}{1-x}$

$\Rightarrow$

$1+x+x^2+x^3+\cdots=\frac{1}{1-x}$

$x+x^2+x^3+\cdots=\boxed{\frac{1}{1-x}-1}$

$\Rightarrow$

$\frac{1}{1-x}-1=999$

$1000-1000x=1$

$999=1000x$

$\therefore \boxed{x=\frac{999}{1000}}$

simply a geometric sequence