If the coefficients of the rth, (r +1)th and (r + 2)th terms in the binomial expansion of (1 + y)^(m)` are in A.P., then m and r satisfy the equation

Options

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

Answer

To solve the problem, we need to find the relationship between \( m \) and \( r \) given that the coefficients of the \( r \)-th, \( (r + 1) \)-th, and \( (r + 2) \)-th terms in the binomial expansion of \( (1 + y)^m \) are in Arithmetic Progression (A.P.).

### Step-by-Step Solution:

1. **Identify the Terms**:
The \( r \)-th term in the binomial expansion of \( (1 + y)^m \) is given by:
\[ T_r = \binom{m}{r} y^r \]
The \( (r + 1) \)-th term is:
\[ T_{r+1} = \binom{m}{r + 1} y^{r + 1} \]
The \( (r + 2) \)-th term is:
\[ T_{r + 2} = \binom{m}{r + 2} y^{r + 2} \]

2. **Set Up the A.P. Condition**:
Since the coefficients of these terms are in A.P., we can express this condition mathematically:
\[ 2 \cdot \binom{m}{r + 1} = \binom{m}{r} + \binom{m}{r + 2} \]

3. **Use the Property of Binomial Coefficients**:
We can use the property of binomial coefficients:
\[ \binom{m}{r + 1} = \frac{m - r}{r + 1} \cdot \binom{m}{r} \]
\[ \binom{m}{r + 2} = \frac{m - r - 1}{r + 2} \cdot \binom{m}{r + 1} \]
Substituting these into the A.P. condition gives:
\[ 2 \cdot \frac{m - r}{r + 1} \cdot \binom{m}{r} = \binom{m}{r} + \frac{m - r - 1}{r + 2} \cdot \frac{m - r}{r + 1} \cdot \binom{m}{r} \]

4. **Simplify the Equation**:
Dividing through by \( \binom{m}{r} \) (assuming \( \binom{m}{r} \neq 0 \)):
\[ 2 \cdot \frac{m - r}{r + 1} = 1 + \frac{(m - r - 1)(m - r)}{(r + 2)(r + 1)} \]
Cross-multiplying and simplifying leads to:
\[ 2(m - r)(r + 2) = (r + 1) + (m - r - 1)(m - r) \]

5. **Rearranging the Equation**:
Rearranging and simplifying gives us a quadratic equation in \( m \):
\[ m^2 - m(4r + 1) + 4r^2 - 2 = 0 \]

6. **Final Result**:
Thus, we conclude that \( m \) and \( r \) satisfy the equation:
\[ m^2 - m(4r + 1) + 4r^2 - 2 = 0 \]