If the first three terms in the expansion of `(1 + ax)^(n)` in ascending powers of `x` are `1+ 12x + 64x^(2)`, find `n and a`.

Answer

To solve the problem, we need to find the values of \( n \) and \( a \) given that the first three terms in the expansion of \( (1 + ax)^n \) are \( 1 + 12x + 64x^2 \).

### Step 1: Write the Binomial Expansion
The binomial expansion of \( (1 + ax)^n \) can be expressed as:
\[ (1 + ax)^n = \sum_{k=0}^{n} \binom{n}{k} (ax)^k \]
The first three terms of this expansion are:
\[ \binom{n}{0} (1) + \binom{n}{1} (ax) + \binom{n}{2} (ax)^2 \]
This simplifies to:
\[ 1 + n(ax) + \frac{n(n-1)}{2}(a^2x^2) \]
Thus, the first three terms are:
\[ 1 + nax + \frac{n(n-1)}{2}a^2x^2 \]

### Step 2: Set Up the Equations
From the problem statement, we know:
\[ 1 + nax + \frac{n(n-1)}{2}a^2x^2 = 1 + 12x + 64x^2 \]
By comparing coefficients, we can form the following equations:
1. \( na = 12 \) (coefficient of \( x \))
2. \( \frac{n(n-1)}{2}a^2 = 64 \) (coefficient of \( x^2 \))

### Step 3: Solve for \( a \)
From the first equation, we can express \( a \) in terms of \( n \):
\[ a = \frac{12}{n} \]

### Step 4: Substitute \( a \) into the Second Equation
Substituting \( a \) into the second equation:
\[ \frac{n(n-1)}{2}\left(\frac{12}{n}\right)^2 = 64 \]
This simplifies to:
\[ \frac{n(n-1)}{2} \cdot \frac{144}{n^2} = 64 \]
Multiplying both sides by \( 2n^2 \):
\[ n(n-1) \cdot 144 = 128n^2 \]
Rearranging gives:
\[ 144n(n-1) = 128n^2 \]
\[ 144n^2 - 144n = 128n^2 \]
\[ 16n^2 - 144n = 0 \]

### Step 5: Factor the Equation
Factoring out \( 16n \):
\[ 16n(n - 9) = 0 \]
This gives us two solutions:
1. \( n = 0 \)
2. \( n = 9 \)

Since \( n \) must be a positive integer, we take \( n = 9 \).

### Step 6: Find \( a \)
Now substituting \( n = 9 \) back into the equation for \( a \):
\[ a = \frac{12}{9} = \frac{4}{3} \]

### Conclusion
Thus, the values of \( n \) and \( a \) are:
\[ n = 9, \quad a = \frac{4}{3} \]