$$  $$In the expansion of $$ \left(3-2x\right)\left(1+\frac{x}{2}\right)^{n} $$. The coefficient of x is 7. Find the value of the constant n and hence find the coefficient of $$ x^2 $$

We are tasked with expanding $ \left(3 - 2x\right)\left(1 + \frac{x}{2}\right)^{n} $ and finding the value of $ n $ such that the coefficient of $ x $ is 7. Afterward, we will also determine the coefficient of $ x^2 $.

### Step 1: Expand $ \left(1 + \frac{x}{2}\right)^{n} $

We will use the binomial theorem to expand $ \left(1 + \frac{x}{2}\right)^{n} $ up to the terms involving $ x $ and $ x^2 $, since we're interested in the coefficients of $ x $ and $ x^2 $. The binomial expansion is:

\[
\left(1 + \frac{x}{2}\right)^{n} = 1 + \binom{n}{1} \left(\frac{x}{2}\right) + \binom{n}{2} \left(\frac{x}{2}\right)^2 + \cdots
\]

This simplifies to:

\[
\left(1 + \frac{x}{2}\right)^{n} = 1 + n\frac{x}{2} + \frac{n(n-1)}{2} \left(\frac{x}{2}\right)^2 + \cdots
\]

\[
\left(1 + \frac{x}{2}\right)^{n} = 1 + \frac{n}{2}x + \frac{n(n-1)}{8}x^2 + \cdots
\]

### Step 2: Multiply by $ 3 - 2x $

Now, we multiply the expansion of $ \left(1 + \frac{x}{2}\right)^{n} $ by $ 3 - 2x $:

\[
(3 - 2x)\left(1 + \frac{n}{2}x + \frac{n(n-1)}{8}x^2 + \cdots\right)
\]

Distribute $ 3 - 2x $:

\[
= 3(1 + \frac{n}{2}x + \frac{n(n-1)}{8}x^2 + \cdots) - 2x(1 + \frac{n}{2}x + \frac{n(n-1)}{8}x^2 + \cdots)
\]

This simplifies to:

\[
= 3 + \frac{3n}{2}x + \frac{3n(n-1)}{8}x^2 - 2x - n x^2 + \cdots
\]

### Step 3: Collect terms involving $ x $

From the expansion, the terms involving $ x $ are:

\[
\frac{3n}{2}x - 2x
\]

For the coefficient of $ x $ to be 7, we set up the equation:

\[
\frac{3n}{2} - 2 = 7
\]

Solve for $ n $:

\[
\frac{3n}{2} = 9
\]

\[
3n = 18
\]

\[
n = 6
\]

### Step 4: Find the coefficient of $ x^2 $

Now that we know $ n = 6 $, substitute this into the terms involving $ x^2 $. The terms involving $ x^2 $ are:

\[
\frac{3n(n-1)}{8}x^2 - n x^2
\]

Substitute $ n = 6 $:

\[
\frac{3 \times 6 \times 5}{8}x^2 - 6x^2
\]

Simplify:

\[
\frac{90}{8}x^2 - 6x^2 = 11.25x^2 - 6x^2 = 5.25x^2
\]

Thus, the coefficient of $ x^2 $ is $ 5.25 $.

### Final Answer:
- The value of $ n $ is $ 6 $.
- The coefficient of $ x^2 $ is $ 5.25 $.

Would you like further details on any step?

Here are five related questions for practice:
1. What is the binomial expansion of $ \left(1 + \frac{x}{3}\right)^5 $?
2. How would the expansion change if the expression was $ (2 - x)\left(1 + \frac{x}{3}\right)^n $?
3. What is the general form of the binomial expansion for $ (a + bx)^n $?
4. How do you find the coefficient of a specific power in a binomial expansion?
5. What are the properties of binomial coefficients?

**Tip**: In binomial expansions, identifying the terms you need early on can save a lot of time.

In the expansion of (3 - 2x)(1 + x/2)^n, the coefficient of x is 7. Find the value of the constant n and hence find the coefficient of x^2.

This problem involves expanding (3 - 2x)(1 + x/2)^n using the binomial theorem and finding the value of n such that the coefficient of x is 7. Then, the coefficient of x^2 is calculated. This solution is suitable for students in Grades 10-12.

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