To solve question 8b, we first need to analyze the binomial expansion of $(1 + ax)^{10}$.

Step 1: Binomial Expansion for First 4 Terms

The binomial expansion of $(1 + ax)^{10}$ can be expressed as:

$(1 + ax)^{10} = \sum_{k=0}^{10} \binom{10}{k} (1)^{10-k} (ax)^k$

We need the first 4 terms of the expansion in ascending powers of $x$. Therefore, we calculate:

1. Term 1 (k = 0): $\binom{10}{0} (1)^10 (ax)^0 = 1$

2. Term 2 (k = 1): $\binom{10}{1} (1)^9 (ax)^1 = 10(ax) = 10ax$

3. Term 3 (k = 2): $\binom{10}{2} (1)^8 (ax)^2 = 45a^2x^2$

4. Term 4 (k = 3): $\binom{10}{3} (1)^7 (ax)^3 = 120a^3x^3$

Thus, the first 4 terms of the expansion are: $1 + 10ax + 45a^2x^2 + 120a^3x^3$

Step 2: Set Up the Condition from Question 8b

We are given that the coefficient of $x^3$ is double the coefficient of $x^2$.

Coefficient of $x^2$: $45a^2$
Coefficient of $x^3$: $120a^3$

According to the condition: $120a^3 = 2 \times 45a^2$

Step 3: Solve for $a$

Simplify the equation: $120a^3 = 90a^2$

Divide both sides by $30a^2$ (assuming $a \neq 0$): $4a = 3$

$a = \frac{3}{4}$

Final Answer

The value of $a$ is $\frac{3}{4}$.

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Would you like more details, or do you have any further questions? Here are some related questions you might find interesting:

1. What is the binomial expansion formula, and how is it derived?
2. How do you find a specific term in a binomial expansion?
3. Can you generalize this problem to higher powers or different binomials?
4. How does changing the coefficient in the binomial expression affect the expansion?
5. What are the applications of binomial expansions in real-world problems?

Tip: Remember that the binomial theorem is a powerful tool for expanding expressions raised to a power, and it's often used in probability, statistics, and algebraic manipulations!