We are tasked with finding the coefficient of $x$ in the expansion of:

$(1 + \frac{3}{x})^2 \left(1 + \frac{3x}{4}\right)^6$

Step 1: Expand $(1 + \frac{3}{x})^2$

Using the binomial expansion, we get:

$(1 + \frac{3}{x})^2 = 1 + 2 \cdot \frac{3}{x} + \left(\frac{3}{x}\right)^2 = 1 + \frac{6}{x} + \frac{9}{x^2}$

Step 2: Expand $\left(1 + \frac{3x}{4}\right)^6$

Using the binomial expansion again, we have:

$\left(1 + \frac{3x}{4}\right)^6 = \sum_{k=0}^6 \binom{6}{k} \left(\frac{3x}{4}\right)^k$

The first few terms are:

$1 + 6\left(\frac{3x}{4}\right) + 15\left(\frac{3x}{4}\right)^2 + 20\left(\frac{3x}{4}\right)^3 + \cdots$ Simplifying each term: $= 1 + \frac{18x}{4} + \frac{135x^2}{16} + \frac{540x^3}{64} + \cdots$ $= 1 + \frac{9x}{2} + \frac{135x^2}{16} + \frac{135x^3}{16} + \cdots$

Step 3: Multiply the two expansions

Now, multiply $(1 + \frac{6}{x} + \frac{9}{x^2})$ by $\left(1 + \frac{9x}{2} + \frac{135x^2}{16} + \cdots\right)$.

We are only interested in the terms that produce $x$.

$(1 \cdot \frac{9x}{2}) + \left(\frac{6}{x} \cdot \frac{135x^2}{16}\right)$ Simplifying:

• From $1 \cdot \frac{9x}{2}$, we get $\frac{9x}{2}$.
• From $\frac{6}{x} \cdot \frac{135x^2}{16}$, we get $\frac{810x}{16} = \frac{405x}{8}$.

Step 4: Add the two results

$\frac{9x}{2} + \frac{405x}{8} = \frac{36x}{8} + \frac{405x}{8} = \frac{441x}{8}$

Thus, the coefficient of $x$ is:

$\boxed{\frac{441}{8}}$

Would you like further details on this or have any questions?

Here are some related questions:

1. What is the binomial expansion formula for general cases?
2. How do you find coefficients in binomial expansions?
3. How would the expansion change for different powers?
4. Can you explain why we only focused on certain terms in the product?
5. How do we expand products involving inverse powers?

Tip: When multiplying expansions, always group and simplify terms carefully to avoid errors in coefficients.