### Instruction: If $1+2x+3x^2 + \dotsb=9$, find $x$.### Response: This is an infinite series, so I need to find a formula for the sum in terms of x.
I notice that the terms are increasing powers of x, multiplied by the corresponding natural number.
This reminds me of the formula for the sum of a geometric series: $a+ar+ar^2 + \dotsb = \frac{a}{1-r}$, where $a$ is the first term and $r$ is the common ratio.
In this case, $a=1$ and $r=x$, so the formula becomes $1+x+x^2 + \dotsb = \frac{1}{1-x}$.
But I also have the extra factor of $n$ in each term, where $n$ is the exponent of $x$.
To account for this, I can try differentiating both sides of the equation with respect to $x$.
On the left side, I get $1+2x+3x^2 + \dotsb$, which is exactly what I want.
On the right side, I get $\frac{d}{dx} \frac{1}{1-x} = \frac{1}{(1-x)^2}$, by using the quotient rule.
So I have $1+2x+3x^2 + \dotsb = \frac{1}{(1-x)^2}$.
Now I can use the given value of the sum, which is $9$, and solve for $x$.
I have $\frac{1}{(1-x)^2} = 9$, which implies that $(1-x)^2 = \frac{1}{9}$.
Taking square roots of both sides, I get $1-x = \pm \frac{1}{3}$.
Adding $x$ to both sides, I get $x = 1 \mp \frac{1}{3}$.
Therefore, $x = \frac{2}{3}$ or $x = \frac{4}{3}$.