If the coefficients of 2nd, 3rd and 4th terms in the expansion of `(1+x)^n \ , nin NN` are in A.P, then `n` is

Options

A. `7`
B. `14`
C. `2`
D. none of these

Explanation

To solve the problem, we need to find the value of \( n \) such that the coefficients of the 2nd, 3rd, and 4th terms in the expansion of \( (1 + x)^n \) are in Arithmetic Progression (A.P.).

### Step-by-Step Solution:

1. **Identify the Coefficients**:
The general term in the expansion of \( (1 + x)^n \) is given by:
\[ T_{r+1} = \binom{n}{r} x^r \]
Therefore, the coefficients of the 2nd, 3rd, and 4th terms are:
- Coefficient of 2nd term: \( \binom{n}{1} \)
- Coefficient of 3rd term: \( \binom{n}{2} \)
- Coefficient of 4th term: \( \binom{n}{3} \)

2. **Set Up the A.P. Condition**:
For the coefficients to be in A.P., the condition is:
\[ 2 \cdot \binom{n}{2} = \binom{n}{1} + \binom{n}{3} \]

3. **Substituting the Binomial Coefficients**:
Substitute the values of the binomial coefficients:
- \( \binom{n}{1} = n \)
- \( \binom{n}{2} = \frac{n(n-1)}{2} \)
- \( \binom{n}{3} = \frac{n(n-1)(n-2)}{6} \)

Therefore, the equation becomes:
\[ 2 \cdot \frac{n(n-1)}{2} = n + \frac{n(n-1)(n-2)}{6} \]

4. **Simplifying the Equation**:
Simplifying the left side:
\[ n(n-1) = n + \frac{n(n-1)(n-2)}{6} \]
Multiply the entire equation by 6 to eliminate the fraction:
\[ 6n(n-1) = 6n + n(n-1)(n-2) \]

5. **Rearranging the Equation**:
Rearranging gives:
\[ 6n^2 - 6n = 6n + n^3 - 3n^2 + 2n \]
Combine like terms:
\[ 6n^2 - 6n - 6n - 2n = n^3 - 3n^2 \]
This simplifies to:
\[ 0 = n^3 - 9n^2 + 0 \]

6. **Factoring the Equation**:
Factor out \( n^2 \):
\[ n^2(n - 9) = 0 \]
This gives us two solutions:
\[ n^2 = 0 \quad \text{or} \quad n - 9 = 0 \]
Thus, \( n = 0 \) or \( n = 9 \).

7. **Considering the Condition**:
Since \( n \) must be a natural number, we discard \( n = 0 \) and accept:
\[ n = 9 \]

### Final Answer:
The value of \( n \) is \( 9 \).