In the expansion of `( 1+ ax)^(n)`, first three terms are 1,12x and `64x^(2)`, then `n =?`

Options

A. 6
B. 9
C. 10
D. 12

Answer

To find the value of \( n \) in the expansion of \( (1 + ax)^n \) where the first three terms are given as \( 1, 12x, \) and \( 64x^2 \), we can follow these steps:

### Step 1: Identify the General Term
The general term in the binomial expansion of \( (1 + ax)^n \) is given by:
\[ T_k = \binom{n}{k} (1)^{n-k} (ax)^k = \binom{n}{k} a^k x^k \]

### Step 2: First Term
The first term (when \( k = 0 \)):
\[ T_0 = \binom{n}{0} a^0 x^0 = 1 \]
This matches the given first term.

### Step 3: Second Term
The second term (when \( k = 1 \)):
\[ T_1 = \binom{n}{1} a^1 x^1 = n \cdot ax \]
We know from the problem that this term equals \( 12x \):
\[ n \cdot a = 12 \quad \text{(1)} \]

### Step 4: Third Term
The third term (when \( k = 2 \)):
\[ T_2 = \binom{n}{2} a^2 x^2 = \frac{n(n-1)}{2} a^2 x^2 \]
We know this term equals \( 64x^2 \):
\[ \frac{n(n-1)}{2} a^2 = 64 \quad \text{(2)} \]

### Step 5: Solve the Equations
From equation (1):
\[ a = \frac{12}{n} \]

Substituting \( a \) into equation (2):
\[ \frac{n(n-1)}{2} \left(\frac{12}{n}\right)^2 = 64 \]
Simplifying this:
\[ \frac{n(n-1)}{2} \cdot \frac{144}{n^2} = 64 \]
\[ \frac{72(n-1)}{n} = 64 \]
Cross-multiplying gives:
\[ 72(n-1) = 64n \]
Expanding this:
\[ 72n - 72 = 64n \]
Rearranging gives:
\[ 72n - 64n = 72 \]
\[ 8n = 72 \]
\[ n = 9 \]

### Conclusion
Thus, the value of \( n \) is:
\[ \boxed{9} \]