If the coefficient of `(2r+1)` th and `(r+2)` th terms in the expansion of `(1+x)^(43)` are equal, then the value of `r(r!=1)` is

Options

A. 5
B. 14
C. 21
D. 22

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

### Step-by-Step Solution:

1. **Understanding the Coefficients**:
The coefficient of the \( k \)th term in the expansion of \( (1 + x)^n \) is given by \( \binom{n}{k-1} \). Here, \( n = 43 \).

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

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

4. **Setting the Coefficients Equal**:
We are given that these coefficients are equal:
\[ \binom{43}{2r} = \binom{43}{r + 1} \]

5. **Using the Property of Binomial Coefficients**:
From the property of binomial coefficients, we know that \( \binom{n}{k} = \binom{n}{n-k} \). Thus, we can set up two equations:
- Case 1: \( 2r = r + 1 \)
- Case 2: \( 2r + (r + 1) = 43 \)

6. **Solving Case 1**:
\[ 2r = r + 1 \implies 2r - r = 1 \implies r = 1 \]
However, we are given that \( r \neq 1 \).

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

8. **Final Answer**:
The value of \( r \) is \( 14 \).