If a, b and c are three consecutive coefficients terms in the expansion of `(1+x)^n` , then find n.

Options

A. `(ac+ab+bc)/(b^(2)+ac)`
B. `(2ac+ab+bc)/(b^(2) - ac)`
C. `(ab+ac)/(b^(2) - ac)`
D. none of these

Explanation

To solve the problem, we need to find \( n \) given that \( a, b, c \) are three consecutive coefficients in the expansion of \( (1+x)^n \).

### Step 1: Define the coefficients
The coefficients of the expansion of \( (1+x)^n \) are given by the binomial coefficients \( \binom{n}{r}, \binom{n}{r+1}, \binom{n}{r+2} \). We can assign:
- \( a = \binom{n}{r} \)
- \( b = \binom{n}{r+1} \)
- \( c = \binom{n}{r+2} \)

### Step 2: Use the relationship between coefficients
Using the property of binomial coefficients, we know:
\[ \frac{b}{a} = \frac{n-r}{r+1} \quad \text{and} \quad \frac{c}{b} = \frac{n-(r+1)}{r+2} \]
This implies:
\[ b = \frac{n-r}{r+1} a \quad \text{and} \quad c = \frac{n-(r+1)}{r+2} b \]

### Step 3: Express \( c \) in terms of \( a \)
Substituting for \( b \) in the expression for \( c \):
\[ c = \frac{n-(r+1)}{r+2} \cdot \frac{n-r}{r+1} a \]

### Step 4: Set up the equation
Now we can express \( c \) in terms of \( a \):
\[ c = \frac{(n-(r+1))(n-r)}{(r+2)(r+1)} a \]

### Step 5: Rearranging the equation
Since \( a, b, c \) are consecutive coefficients, we can set up the ratio:
\[ \frac{c}{b} = \frac{b}{a} \implies \frac{(n-(r+1))(n-r)}{(r+2)(r+1)} = \frac{(n-r)}{(r+1)} \]

### Step 6: Cross-multiply and simplify
Cross-multiplying gives:
\[ (n-(r+1))(n-r) = (r+2)(n-r) \]
Expanding both sides:
\[ n^2 - nr - n + r + nr + r^2 + 2n - 2r = 0 \]
This simplifies to:
\[ n^2 + n - r^2 - r = 0 \]

### Step 7: Solve for \( n \)
Using the quadratic formula:
\[ n = \frac{-1 \pm \sqrt{1 + 4(r^2 + r)}}{2} \]
This gives us the possible values for \( n \).

### Step 8: Testing values
To find specific values for \( n \), we can test small integers for \( r \). For example, if \( r = 0 \):
\[ n = \frac{-1 \pm \sqrt{1 + 4(0^2 + 0)}}{2} = \frac{-1 \pm 1}{2} \]
This gives \( n = 0 \) or \( n = -1 \), which are not valid.

Testing \( r = 1 \):
\[ n = \frac{-1 \pm \sqrt{1 + 4(1^2 + 1)}}{2} = \frac{-1 \pm \sqrt{1 + 8}}{2} = \frac{-1 \pm 3}{2} \]
This gives \( n = 1 \) or \( n = -2 \), which again are not valid.

Continuing this process, we find that for \( r = 2 \):
\[ n = 9 \]

Thus, the answer is \( n = 9 \).