Let's work through the questions from the image.

Part (a)

We are asked to find the coefficient of $x^6y^2$ in the expansion of $(3x^2 + y)^5$ using the binomial theorem.

The binomial theorem states that: $(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$ For the expansion of $(3x^2 + y)^5$:

• $a = 3x^2$,
• $b = y$,
• $n = 5$.

To find the coefficient of $x^6y^2$, we need to choose terms such that the powers of $x$ and $y$ match $x^6y^2$. Notice:

• The term $x^6$ comes from $(3x^2)^3 = 27x^6$,
• The term $y^2$ comes from $y^2$.

This means we need the term in the binomial expansion where $k = 2$ (so that we get $y^2$). The general term in the expansion is: $\binom{5}{k} (3x^2)^{5-k} y^k$ For $k = 2$, this becomes: $\binom{5}{2} (3x^2)^{3} y^2 = \binom{5}{2} \cdot 27x^6 \cdot y^2$ Now, calculate the binomial coefficient: $\binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10$ Thus, the coefficient of $x^6y^2$ is: $10 \cdot 27 = 270$ So, the coefficient of $x^6y^2$ in the expansion of $(3x^2 + y)^5$ is 270.

Part (b)

We are asked to interpret: $[x_1x_2x_3^2](x_1 + 2x_2 + 3x_3)^4$ This is a problem in combinatorics. We are looking for the coefficient of the term $x_1x_2x_3^2$ in the expansion of $(x_1 + 2x_2 + 3x_3)^4$.

This expression can be understood as counting how many ways the variables $x_1$, $x_2$, and $x_3$ can be selected to form the term $x_1x_2x_3^2$. The expansion represents a multinomial, and the task is to determine the number of ways to obtain the specific combination of powers $x_1^1x_2^1x_3^2$ when distributing four total factors from $(x_1 + 2x_2 + 3x_3)^4$.

For each selection:

• Picking $x_1$ from the first term contributes 1 to the total count,
• Picking $x_2$ from the second term contributes 2 to the total count,
• Picking $x_3$ from the third term contributes 3 to the total count.

To form the desired term $x_1x_2x_3^2$, the multinomial coefficient is given by: $\text{Coefficient of } x_1x_2x_3^2 = \frac{4!}{1!1!2!} \cdot (1^1)(2^1)(3^2)$ We will calculate this:

$\text{Coefficient} = \frac{4!}{1!1!2!} \cdot 1 \cdot 2 \cdot 9 = \frac{24}{2} \cdot 18 = 12 \cdot 18 = 216$

So, the coefficient of $x_1x_2x_3^2$ is 216, which can be interpreted as the number of ways to select the variables $x_1$, $x_2$, and $x_3^2$ in the given multinomial expansion.

Would you like further details or explanations on any part of the solution? Here are five related questions to explore the concepts:

1. How does the binomial theorem apply to non-linear terms like $x^2$?
2. What is the significance of binomial coefficients in expansions?
3. How do multinomial coefficients generalize the binomial theorem?
4. What combinatorial interpretation can be made for the term $x^2y^3$ in a different expansion?
5. Can you compute the coefficient of $x^4$ in $(2x + 3y)^6$?

Tip: Understanding the binomial theorem helps simplify calculations in probability, combinatorics, and algebraic expansions.