In the binomial expansion of `(1+x)^(43)`, the coefficients of the `(2r+1)` th and `(r+2)` th terms are equal. Then r=

Options

A. 14
B. 41
C. 1
D. 42

Answer

To solve the problem, we need to find the value of \( r \) such that the coefficients of the \( (2r + 1) \)th term and the \( (r + 2) \)th term in the binomial expansion of \( (1 + x)^{43} \) are equal.

### Step-by-Step Solution:

1. **Identify the General Term**:
The general term \( T_k \) in the binomial expansion of \( (1 + x)^n \) is given by:
\[ T_k = \binom{n}{k-1} \cdot x^{k-1} \]
For our case, \( n = 43 \).

2. **Find the Coefficient of the \( (2r + 1) \)th Term**:
The \( (2r + 1) \)th term corresponds to \( k = 2r + 1 \):
\[ T_{2r + 1} = \binom{43}{2r} \cdot x^{2r} \]
Thus, the coefficient is \( \binom{43}{2r} \).

3. **Find the Coefficient of the \( (r + 2) \)th Term**:
The \( (r + 2) \)th term corresponds to \( k = r + 2 \):
\[ T_{r + 2} = \binom{43}{r + 1} \cdot x^{r + 1} \]
Thus, the coefficient is \( \binom{43}{r + 1} \).

4. **Set the Coefficients Equal**:
Since the coefficients are equal, we have:
\[ \binom{43}{2r} = \binom{43}{r + 1} \]

5. **Use the Property of Binomial Coefficients**:
The property states that \( \binom{n}{k} = \binom{n}{n-k} \). Therefore, we can set up two equations:
- Case 1: \( 2r = r + 1 \)
- Case 2: \( 2r + (r + 1) = 43 \)

6. **Solve Case 1**:
\[ 2r = r + 1 \implies r = 1 \]

7. **Solve Case 2**:
\[ 2r + r + 1 = 43 \implies 3r + 1 = 43 \implies 3r = 42 \implies r = 14 \]

8. **Conclusion**:
The possible values of \( r \) are \( r = 1 \) and \( r = 14 \).

### Final Answer:
Thus, the values of \( r \) are \( 1 \) and \( 14 \).