If the coefficients of `(2r +1)th and (4r + 5)` th terms is the expansion of `(1+x)^(10)` are equal then `r=?`

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 \( (4r + 5) \)th term in the expansion of \( (1+x)^{10} \) 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:
\[ T_k = \binom{n}{k-1} x^{k-1} \]
For \( n = 10 \), the general term becomes:
\[ T_k = \binom{10}{k-1} x^{k-1} \]

2. **Find the Coefficient of the \( (2r + 1) \)th Term**:
The \( (2r + 1) \)th term corresponds to \( k = 2r + 1 \). Thus, the coefficient is:
\[ \text{Coefficient of } T_{2r + 1} = \binom{10}{(2r + 1) - 1} = \binom{10}{2r} \]

3. **Find the Coefficient of the \( (4r + 5) \)th Term**:
The \( (4r + 5) \)th term corresponds to \( k = 4r + 5 \). Thus, the coefficient is:
\[ \text{Coefficient of } T_{4r + 5} = \binom{10}{(4r + 5) - 1} = \binom{10}{4r + 4} \]

4. **Set the Coefficients Equal**:
Since the coefficients are equal, we have:
\[ \binom{10}{2r} = \binom{10}{4r + 4} \]

5. **Use the Property of Binomial Coefficients**:
We know that \( \binom{n}{k} = \binom{n}{n-k} \). Therefore:
\[ \binom{10}{4r + 4} = \binom{10}{10 - (4r + 4)} = \binom{10}{6 - 4r} \]
Thus, we can set:
\[ 2r = 6 - 4r \]

6. **Solve for \( r \)**:
Rearranging the equation gives:
\[ 2r + 4r = 6 \implies 6r = 6 \implies r = 1 \]

### Final Answer:
The value of \( r \) is:
\[ \boxed{1} \]