if the coefficient of `(2r+1)`th term and `(r+2)`th term in the expansion of `(1+x)^(43)` are equal then r=?

Options

A. 14
B. 30
C. 41
D. 42

Explanation

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 expansion of \( (1 + x)^{43} \) are equal.

### Step-by-Step Solution:

1. **Identify the General Term**:
The general term in the expansion of \( (1 + x)^{43} \) is given by:
\[ T_k = \binom{43}{k} x^k \]
where \( k \) is the term number starting from 0.

2. **Find the Coefficient of the \( (2r + 1) \)th Term**:
The \( (2r + 1) \)th term corresponds to \( k = 2r \) (since the term number starts from 0). Thus, the coefficient is:
\[ \text{Coefficient of } T_{2r + 1} = \binom{43}{2r} \]

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

4. **Set the Coefficients Equal**:
According to the problem, these coefficients are equal:
\[ \binom{43}{2r} = \binom{43}{r + 1} \]

5. **Using the Property of Binomial Coefficients**:
The property of binomial coefficients states that:
\[ \binom{n}{k} = \binom{n}{n-k} \]
Therefore, we can write:
\[ \binom{43}{2r} = \binom{43}{43 - 2r} \]
This gives us two cases to consider:
- Case 1: \( 2r = r + 1 \)
- Case 2: \( 2r = 43 - (r + 1) \)

6. **Solve Case 1**:
From \( 2r = r + 1 \):
\[ 2r - r = 1 \implies r = 1 \]

7. **Solve Case 2**:
From \( 2r = 43 - r - 1 \):
\[ 2r + r = 42 \implies 3r = 42 \implies r = \frac{42}{3} = 14 \]

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

### Final Answer:
Thus, the values of \( r \) that satisfy the condition are:
\[ r = 1 \text{ or } r = 14 \]