If `(1 + ax)^n= 1 +8x+ 24x^2 +.....` then the value of a and n is

Answer

To solve the equation \( (1 + ax)^n = 1 + 8x + 24x^2 + \ldots \), we will follow these steps:

### Step 1: Identify the coefficients
We start with the binomial expansion of \( (1 + ax)^n \):
\[ (1 + ax)^n = \sum_{k=0}^{n} \binom{n}{k} (ax)^k = 1 + \binom{n}{1} ax + \binom{n}{2} (ax)^2 + \ldots \]
This can be rewritten as:
\[ 1 + n a x + \frac{n(n-1)}{2} a^2 x^2 + \ldots \]

### Step 2: Compare coefficients
From the given expression \( 1 + 8x + 24x^2 + \ldots \), we can compare coefficients:
- Coefficient of \( x \):
\[ na = 8 \quad \text{(1)} \]
- Coefficient of \( x^2 \):
\[ \frac{n(n-1)}{2} a^2 = 24 \quad \text{(2)} \]

### Step 3: Solve for \( a \) from equation (1)
From equation (1), we can express \( a \) in terms of \( n \):
\[ a = \frac{8}{n} \quad \text{(3)} \]

### Step 4: Substitute \( a \) into equation (2)
Now we substitute equation (3) into equation (2):
\[ \frac{n(n-1)}{2} \left(\frac{8}{n}\right)^2 = 24 \]
This simplifies to:
\[ \frac{n(n-1)}{2} \cdot \frac{64}{n^2} = 24 \]
Multiplying both sides by \( 2n^2 \) gives:
\[ 64(n-1) = 48n \]
Expanding and rearranging:
\[ 64n - 64 = 48n \]
\[ 64n - 48n = 64 \]
\[ 16n = 64 \]
\[ n = 4 \quad \text{(4)} \]

### Step 5: Substitute \( n \) back to find \( a \)
Now we substitute \( n = 4 \) back into equation (3) to find \( a \):
\[ a = \frac{8}{4} = 2 \quad \text{(5)} \]

### Conclusion
Thus, the values of \( a \) and \( n \) are:
\[ a = 2, \quad n = 4 \]