If the coefficient of second, third and fourth terms in the expansion of `(1+x)^(n)` are in AP, then n is equal to

Options

A. 7
B. 4
C. 5
D. 6

Answer

To solve the problem, we need to find the value of \( n \) such that the coefficients of the second, third, and fourth terms in the expansion of \( (1+x)^n \) are in arithmetic progression (AP).

### Step-by-Step Solution:

1. **Identify the Coefficients**:
The general term in the binomial expansion of \( (1+x)^n \) is given by:
\[ T_{r+1} = \binom{n}{r} x^r \]
Therefore, the coefficients of the second, third, and fourth terms are:
- Second term: \( T_2 = \binom{n}{1} = n \)
- Third term: \( T_3 = \binom{n}{2} = \frac{n(n-1)}{2} \)
- Fourth term: \( T_4 = \binom{n}{3} = \frac{n(n-1)(n-2)}{6} \)

2. **Set Up the Arithmetic Progression Condition**:
For the coefficients to be in AP, the condition is:
\[ 2 \cdot \binom{n}{2} = \binom{n}{1} + \binom{n}{3} \]
Substituting the coefficients we found:
\[ 2 \cdot \frac{n(n-1)}{2} = n + \frac{n(n-1)(n-2)}{6} \]
Simplifying the left side:
\[ n(n-1) = n + \frac{n(n-1)(n-2)}{6} \]

3. **Multiply Through by 6 to Eliminate the Denominator**:
\[ 6n(n-1) = 6n + n(n-1)(n-2) \]

4. **Expand and Rearrange**:
Expanding the right side:
\[ 6n(n-1) = 6n + n(n^2 - 3n + 2) \]
This simplifies to:
\[ 6n(n-1) = 6n + n^3 - 3n^2 + 2n \]
Rearranging gives:
\[ n^3 - 9n^2 + 8n = 0 \]

5. **Factor the Polynomial**:
Factor out \( n \):
\[ n(n^2 - 9n + 8) = 0 \]
This gives us:
\[ n = 0 \quad \text{or} \quad n^2 - 9n + 8 = 0 \]

6. **Solve the Quadratic Equation**:
Using the quadratic formula:
\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{9 \pm \sqrt{81 - 32}}{2} = \frac{9 \pm \sqrt{49}}{2} = \frac{9 \pm 7}{2} \]
This gives:
\[ n = \frac{16}{2} = 8 \quad \text{or} \quad n = \frac{2}{2} = 1 \]

7. **Check for Validity**:
We check if \( n = 1 \) is valid:
- If \( n = 1 \), the coefficients are \( 1, 0, 0 \) which do not form an AP.
Thus, the only valid solution is:
\[ n = 8 \]

### Final Answer:
\[ n = 8 \]