To determine the values of [tex]\( n \)[/tex] and [tex]\( a \)[/tex] given the first three terms of the expansion of [tex]\( (1 + ax)^n \)[/tex], we start with the binomial expansion formula:
[tex]\[
(1 + ax)^n = \sum_{k=0}^{n} \binom{n}{k} (ax)^k
\][/tex]
The first three terms of this expansion are:
[tex]\[
\binom{n}{0} (ax)^0 = 1, \quad \binom{n}{1} (ax)^1 = nax, \quad \binom{n}{2} (ax)^2 = \frac{n(n-1)}{2} a^2 x^2
\][/tex]
Given that the first three terms of the expansion are [tex]\( 1 + 12x + 64x^2 \)[/tex], we can match these coefficients with our expansion:
1. For the constant term:
[tex]\[
\binom{n}{0} (ax)^0 = 1 \quad \Rightarrow \quad 1 = 1
\][/tex]
2. For the coefficient of [tex]\( x \)[/tex]:
[tex]\[
\binom{n}{1} (ax)^1 = nax \quad \Rightarrow \quad na = 12
\][/tex]
3. For the coefficient of [tex]\( x^2 \)[/tex]:
[tex]\[
\binom{n}{2} (ax)^2 = \frac{n(n-1)}{2} a^2 x^2 \quad \Rightarrow \quad \frac{n(n-1)}{2} a^2 = 64
\][/tex]
Now, solve these equations step-by-step.
First, from [tex]\( na = 12 \)[/tex], we solve for [tex]\( a \)[/tex]:
[tex]\[
a = \frac{12}{n}
\][/tex]
Next, substitute [tex]\( a \)[/tex] into the equation for the [tex]\( x^2 \)[/tex] term:
[tex]\[
\frac{n(n-1)}{2} \left( \frac{12}{n} \right)^2 = 64
\][/tex]
Simplify the equation:
[tex]\[
\frac{n(n-1)}{2} \cdot \frac{144}{n^2} = 64 \quad \Rightarrow \quad \frac{144(n-1)}{2n} = 64 \quad \Rightarrow \quad \frac{72(n-1)}{n} = 64
\][/tex]
Multiply both sides by [tex]\( n \)[/tex]:
[tex]\[
72(n-1) = 64n \quad \Rightarrow \quad 72n - 72 = 64n \quad \Rightarrow \quad 8n = 72 \quad \Rightarrow \quad n = 9
\][/tex]
Now, substitute [tex]\( n = 9 \)[/tex] back into the equation [tex]\( na = 12 \)[/tex]:
[tex]\[
9a = 12 \quad \Rightarrow \quad a = \frac{12}{9} = \frac{4}{3}
\][/tex]
Therefore, the values of [tex]\( n \)[/tex] and [tex]\( a \)[/tex] are:
[tex]\[
n = 9 \quad \text{and} \quad a = \frac{4}{3}
\][/tex]
