If `(1 +px)^(n) = 1 + 24x + 264x^(2)` + .. . then

Options

A. p = 2, n = 6
B. p = 2, n = 12
C. p = 3, n = 6
D. none of these

Answer

To solve the problem, we need to find the values of \( n \) and \( p \) given the equation:

\[ (1 + px)^n = 1 + 24x + 264x^2 + \ldots \]

### Step 1: Identify the series expansion
We know that the binomial expansion of \( (1 + x)^n \) is given by:

\[ (1 + x)^n = \sum_{k=0}^{n} \binom{n}{k} x^k \]

In our case, we replace \( x \) with \( px \):

\[ (1 + px)^n = \sum_{k=0}^{n} \binom{n}{k} (px)^k = \sum_{k=0}^{n} \binom{n}{k} p^k x^k \]

### Step 2: Compare coefficients
From the given series \( 1 + 24x + 264x^2 + \ldots \), we can identify the coefficients:

- The coefficient of \( x^1 \) is \( 24 \).
- The coefficient of \( x^2 \) is \( 264 \).

From the binomial expansion, we have:

- Coefficient of \( x^1 \): \( \binom{n}{1} p = np \)
- Coefficient of \( x^2 \): \( \binom{n}{2} p^2 = \frac{n(n-1)}{2} p^2 \)

### Step 3: Set up equations
From the coefficients, we can set up the following equations:

1. \( np = 24 \) (1)
2. \( \frac{n(n-1)}{2} p^2 = 264 \) (2)

### Step 4: Solve for \( p \) from equation (1)
From equation (1):

\[ p = \frac{24}{n} \]

### Step 5: Substitute \( p \) into equation (2)
Substituting \( p \) into equation (2):

\[ \frac{n(n-1)}{2} \left(\frac{24}{n}\right)^2 = 264 \]

This simplifies to:

\[ \frac{n(n-1)}{2} \cdot \frac{576}{n^2} = 264 \]

### Step 6: Simplify the equation
Multiplying both sides by \( 2n^2 \):

\[ 576n(n-1) = 528n^2 \]

Rearranging gives:

\[ 576n^2 - 576n = 528n^2 \]

This simplifies to:

\[ 48n^2 - 576n = 0 \]

### Step 7: Factor the equation
Factoring out \( 48n \):

\[ 48n(n - 12) = 0 \]

Thus, \( n = 0 \) or \( n = 12 \).

### Step 8: Find \( p \)
Since \( n = 0 \) is not valid in this context, we take \( n = 12 \). Now substituting back to find \( p \):

\[ p = \frac{24}{12} = 2 \]

### Final Answer
Thus, the values of \( n \) and \( p \) are:

\[ n = 12, \quad p = 2 \]