If the coefficients of rth, (r + 1)th and (r +2)th terms in the expansion `(1+ x)^(n)` are in A, P., then

Options

A. `n^(2)+n(4r+1)+4r^(2)-2=0`
B. `n^(2)+n(4r+1)+4r^(2)+2=0`
C. `(n-2r)^(2)=n+2`
D. `(n+2r)^(2)=n+2`

Answer

To solve the problem, we need to find the conditions under which the coefficients of the r-th, (r + 1)-th, and (r + 2)-th terms in the expansion of \( (1 + x)^n \) are in Arithmetic Progression (AP).

### Step 1: Identify the coefficients

The coefficients of the r-th, (r + 1)-th, and (r + 2)-th terms in the expansion of \( (1 + x)^n \) can be expressed using the binomial coefficient:

- Coefficient of the r-th term: \( C_r = \binom{n}{r} \)
- Coefficient of the (r + 1)-th term: \( C_{r+1} = \binom{n}{r + 1} \)
- Coefficient of the (r + 2)-th term: \( C_{r+2} = \binom{n}{r + 2} \)

### Step 2: Set up the condition for AP

The coefficients are in AP if:

\[ 2C_{r + 1} = C_r + C_{r + 2} \]

### Step 3: Substitute the binomial coefficients

Substituting the expressions for the coefficients, we have:

\[ 2 \binom{n}{r + 1} = \binom{n}{r} + \binom{n}{r + 2} \]

### Step 4: Use the property of binomial coefficients

Using the identity for binomial coefficients:

\[ \binom{n}{r + 2} = \binom{n}{r + 1} \cdot \frac{n - (r + 1)}{r + 2} \]

and

\[ \binom{n}{r} = \binom{n}{r + 1} \cdot \frac{r + 1}{n - r} \]

We can rewrite the equation as:

\[ 2 \binom{n}{r + 1} = \binom{n}{r + 1} \cdot \left( \frac{r + 1}{n - r} + \frac{n - (r + 1)}{r + 2} \right) \]

### Step 5: Simplify the equation

Dividing both sides by \( \binom{n}{r + 1} \) (assuming it is not zero), we get:

\[ 2 = \frac{r + 1}{n - r} + \frac{n - (r + 1)}{r + 2} \]

### Step 6: Solve for n

Now, we can solve this equation for \( n \):

1. Cross-multiply to eliminate the fractions.
2. Rearrange the terms to isolate \( n \).
3. Solve for \( n \).

### Final Result

After simplification, we will find the relationship between \( n \) and \( r \) that satisfies the condition for the coefficients to be in AP.

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