In the expansion of `( 1+ x )^(43)`, if the coefficient of `(2r + 1)^(th)` and `( r + 2)^(th)` terms are equal, then what is the value of r `( r cancel(=)1)` ?

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

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

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

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

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

6. **Solve the First Equation**:
From \( 2r = 43 - r - 1 \):
\[ 2r + r + 1 = 43 \implies 3r + 1 = 43 \implies 3r = 42 \implies r = 14 \]

7. **Check the Other Condition**:
The first equation gives \( r = 1 \), which is not allowed as per the problem statement. Thus, we only consider \( r = 14 \).

8. **Conclusion**:
The value of \( r \) is:
\[ \boxed{14} \]