Let's solve the two questions shown in the image one by one:

Question 19:

Given that $\left( 1 + \frac{2}{3}x \right)^n (3 + nx)^2 = 9 + 84x + \dots$, find the value of $n$.

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

We apply the binomial expansion to $\left( 1 + \frac{2}{3}x \right)^n$ up to the first two terms: $\left( 1 + \frac{2}{3}x \right)^n = 1 + n \cdot \frac{2}{3}x + \dots = 1 + \frac{2n}{3}x + \dots$

Step 2: Expand $(3 + nx)^2$:

Expand this quadratic expression: $(3 + nx)^2 = 9 + 2(3)(nx) + (nx)^2 = 9 + 6nx + n^2x^2$

Step 3: Multiply the two expansions:

Multiply the expansions of $\left( 1 + \frac{2}{3}x \right)^n$ and $(3 + nx)^2$: $\left( 1 + \frac{2}{3}x \right)^n (3 + nx)^2 = \left( 1 + \frac{2n}{3}x + \dots \right) \left( 9 + 6nx + n^2x^2 \right)$

First, let's expand the constant term: $9 \times 1 = 9$

Next, for the term with $x$: $9 \times \frac{2n}{3}x + 6nx \times 1 = 6n x + 6n x = 12n x$

We are given that the term with $x$ is $84x$. Therefore, equating the coefficients of $x$, we have: $12n = 84$ $n = \frac{84}{12} = 7$

Thus, the value of $n$ is $\boxed{7}$.

Question 20:

Find the coefficient of $x^{-2}$ in the expansion of $(x - 1)^3 \left( \frac{1}{x} + 2x \right)^6$.

Step 1: Expand $(x - 1)^3$:

We expand $(x - 1)^3$ using the binomial theorem: $(x - 1)^3 = x^3 - 3x^2 + 3x - 1$

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

We apply the binomial expansion to $\left( \frac{1}{x} + 2x \right)^6$: $\left( \frac{1}{x} + 2x \right)^6 = \sum_{k=0}^{6} \binom{6}{k} \left( \frac{1}{x} \right)^{6-k} (2x)^k$ Simplifying each term: $= \binom{6}{0} \frac{1}{x^6} + \binom{6}{1} \frac{1}{x^5} (2x) + \binom{6}{2} \frac{1}{x^4} (2x)^2 + \dots + \binom{6}{6} (2x)^6$

We'll calculate up to the terms involving $x^{-2}$:

• For $k = 2$: $\binom{6}{2} \frac{1}{x^4} (2x)^2 = \binom{6}{2} \frac{1}{x^4} 4x^2 = 15 \cdot 4 \cdot \frac{1}{x^2} = \frac{60}{x^2}$

Step 3: Multiply the two expansions and find the coefficient of $x^{-2}$:

We now multiply the expansions of $(x - 1)^3$ and $\left( \frac{1}{x} + 2x \right)^6$, focusing on terms that will result in $x^{-2}$.

The $x^0$ term from $(x - 1)^3$ (which is $-1$) will multiply with the $x^{-2}$ term from $\left( \frac{1}{x} + 2x \right)^6$, giving: $(-1) \times \frac{60}{x^2} = -60x^{-2}$

Thus, the coefficient of $x^{-2}$ is $\boxed{-60}$.

Let me know if you'd like more details on any part! Here are some related questions for further exploration:

1. How does the binomial theorem work for fractional and negative exponents?
2. Can you explain how higher-order terms in expansions are calculated?
3. What happens when you multiply two binomial expansions?
4. How do we use the concept of combining like terms in expansions?
5. Can the same methods be applied to trinomials or more complex polynomials?

Tip: When working with expansions, always identify which terms are relevant for the problem to avoid unnecessary calculations!