The given equation is:

$\frac{1}{1+x} + \frac{2}{(1+x)^2} + \frac{3}{(1+x)^3} + \dots = 8$

This is an infinite series of the form:

$S = \sum_{n=1}^{\infty} \frac{n}{(1+x)^n}$

To solve this, we can try recognizing this sum as a standard series. This is a variant of the geometric series, and it can be evaluated using some advanced series techniques. Let's go step by step:

Step 1: General form of the series

The series can be written as:

$S = \sum_{n=1}^{\infty} \frac{n}{(1+x)^n}$

Step 2: Recognize the sum

To handle this, recall the sum of a geometric series $\sum_{n=0}^{\infty} r^n = \frac{1}{1-r}$, for $|r| < 1$. We can take a derivative with respect to $r$ to create the series where the terms involve $n$.

The result we are looking for is derived from the following manipulation:

$\frac{d}{dr} \left( \sum_{n=0}^{\infty} r^n \right) = \sum_{n=1}^{\infty} n r^{n-1} = \frac{1}{(1 - r)^2}$

Multiplying both sides by $r$, we obtain:

$\sum_{n=1}^{\infty} n r^n = \frac{r}{(1 - r)^2}$

Step 3: Substitute $r = \frac{1}{1+x}$

In your case, we have $r = \frac{1}{1+x}$, so the sum becomes:

$\sum_{n=1}^{\infty} n \left( \frac{1}{1+x} \right)^n = \frac{\frac{1}{1+x}}{\left( 1 - \frac{1}{1+x} \right)^2}$

Simplifying the denominator:

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

Thus, the sum is:

$\sum_{n=1}^{\infty} n \left( \frac{1}{1+x} \right)^n = \frac{\frac{1}{1+x}}{\left( \frac{x}{1+x} \right)^2} = \frac{1+x}{x^2}$

Step 4: Set the sum equal to 8

Now we can set the sum equal to 8:

$\frac{1+x}{x^2} = 8$

Step 5: Solve for $x$

Multiply both sides by $x^2$:

$1 + x = 8x^2$

Rearranging this equation:

$8x^2 - x - 1 = 0$

Step 6: Solve the quadratic equation

We can solve the quadratic equation $8x^2 - x - 1 = 0$ using the quadratic formula:

$x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(8)(-1)}}{2(8)}$

$x = \frac{1 \pm \sqrt{1 + 32}}{16}$

$x = \frac{1 \pm \sqrt{33}}{16}$

Thus, the two possible solutions for $x$ are:

$x = \frac{1 + \sqrt{33}}{16} \quad \text{or} \quad x = \frac{1 - \sqrt{33}}{16}$

Since $\sqrt{33} \approx 5.74$, the solutions are approximately:

$x \approx \frac{1 + 5.74}{16} \approx 0.429 \quad \text{or} \quad x \approx \frac{1 - 5.74}{16} \approx -0.298$

Final Answer:

$x \approx 0.429 \quad \text{or} \quad x \approx -0.298$

Let me know if you need further details or clarification!

Here are some related questions for further study:

1. How do we derive the sum of a geometric series and apply its general properties?
2. What are other series transformations we can use to solve sums of this form?
3. How can we analyze the convergence of this series for different values of $x$?
4. How can we apply the quadratic formula to non-standard equations in algebra?
5. What are the implications of having two possible roots for $x$ in this type of series?

Tip: When solving series problems, always look for connections to known series formulas (geometric series, power series) to simplify your work.