If the coefficient of `(r+1)` th term and `(r+3)` th term in the expansion of `(1+x)^(20)` are equal, then the value of r is
(i) 8
(ii) 9
(iii) 16
(iv) None of these

Options

A. 8
B. 9
C. 16
D. None of these

Answer

To solve the problem, we need to find the value of \( r \) such that the coefficients of the \( (r+1) \)th term and the \( (r+3) \)th term in the expansion of \( (1+x)^{20} \) 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, \( n = 20 \).

2. **Write the Coefficients for the Required Terms**:
- The coefficient of the \( (r+1) \)th term, which corresponds to \( T_{r} \) (since \( T_k \) starts from \( k=0 \)), is:
\[ \text{Coefficient of } T_{r} = \binom{20}{r} \]
- The coefficient of the \( (r+3) \)th term, which corresponds to \( T_{r+2} \), is:
\[ \text{Coefficient of } T_{r+2} = \binom{20}{r+2} \]

3. **Set the Coefficients Equal**:
According to the problem, these coefficients are equal:
\[ \binom{20}{r} = \binom{20}{r+2} \]

4. **Use the Property of Binomial Coefficients**:
We know that if \( \binom{n}{r} = \binom{n}{p} \), then either:
- \( r = p \) or
- \( r + p = n \)

Applying this to our equation:
- Here, \( r \) corresponds to \( r \) and \( p \) corresponds to \( r + 2 \).
- Thus, we have two cases:
1. \( r = r + 2 \) (which is invalid)
2. \( r + (r + 2) = 20 \)

5. **Solve the Valid Equation**:
From the second case:
\[ 2r + 2 = 20 \]
Simplifying this gives:
\[ 2r = 20 - 2 \]
\[ 2r = 18 \]
\[ r = 9 \]

6. **Conclusion**:
The value of \( r \) is \( 9 \).

### Final Answer:
The value of \( r \) is \( \boxed{9} \).