To represent $5x \ln(1+2x)$ as a power series, we first expand $\ln(1+2x)$ as a power series around $x=0$ using its Taylor series:

$\ln(1+2x) = 2x - \frac{(2x)^2}{2} + \frac{(2x)^3}{3} - \frac{(2x)^4}{4} + \cdots$

Substituting this expansion:

$\ln(1+2x) = 2x - 2x^2 + \frac{8x^3}{3} - \frac{16x^4}{4} + \cdots$

Now, multiply the series by $5x$:

$5x \ln(1+2x) = 5x \left( 2x - 2x^2 + \frac{8x^3}{3} - \frac{16x^4}{4} + \cdots \right)$

Distributing $5x$:

$5x \ln(1+2x) = 10x^2 - 10x^3 + \frac{40x^4}{3} - 20x^5 + \cdots$

Now, we extract the coefficients $c_0, c_1, c_2, c_3, c_4$ of the terms up to $x^4$:

• $c_0 = 0$ (there is no constant term),
• $c_1 = 0$ (there is no $x$-term),
• $c_2 = 10$,
• $c_3 = -10$,
• $c_4 = \frac{40}{3}$.

Thus, the coefficients are:

$c_0 = 0, \quad c_1 = 0, \quad c_2 = 10, \quad c_3 = -10, \quad c_4 = \frac{40}{3}.$

Let me know if you'd like a more detailed explanation!

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5 Questions to Expand:

1. How is the Taylor series for $\ln(1+2x)$ derived?
2. Can the same method apply to $x^2 \ln(1+3x)$?
3. How do we determine the radius of convergence for this series?
4. What is the significance of each coefficient in the power series?
5. How would the power series change if the argument was $\ln(1-2x)$?

Tip:

When expanding logarithmic functions as power series, remember the signs of the terms alternate and the series converges within $|x| < 1/|a|$, where $a$ is the argument multiplier.