If the coefficients of `(r-5)^(t h)` and `(2r-1)^(t h)` terms in the expansion of `(1+x)^(34)` are equal, find `rdot`

Explanation

To solve the problem, we need to find the value of \( r \) such that the coefficients of the \( (r-5)^{th} \) and \( (2r-1)^{th} \) terms in the expansion of \( (1+x)^{34} \) are equal.

### Step-by-step Solution:

1. **Understanding the Coefficient of Terms in Binomial Expansion**:
The coefficient of the \( k^{th} \) term in the expansion of \( (1+x)^n \) is given by \( \binom{n}{k-1} \). Therefore, the coefficient of the \( (r-5)^{th} \) term is:
\[ \text{Coefficient of } (r-5)^{th} = \binom{34}{(r-5)-1} = \binom{34}{r-6} \]
Similarly, the coefficient of the \( (2r-1)^{th} \) term is:
\[ \text{Coefficient of } (2r-1)^{th} = \binom{34}{(2r-1)-1} = \binom{34}{2r-2} \]

2. **Setting the Coefficients Equal**:
Since the coefficients are equal, we can set up the equation:
\[ \binom{34}{r-6} = \binom{34}{2r-2} \]

3. **Using the Property of Binomial Coefficients**:
We can use the property that \( \binom{n}{k} = \binom{n}{n-k} \). Thus, we can rewrite the right-hand side:
\[ \binom{34}{2r-2} = \binom{34}{34 - (2r-2)} = \binom{34}{36 - 2r} \]
Therefore, we have:
\[ \binom{34}{r-6} = \binom{34}{36 - 2r} \]

4. **Setting the Indices Equal**:
From the equality of binomial coefficients, we can equate the indices:
\[ r - 6 = 36 - 2r \]

5. **Solving for \( r \)**:
Rearranging the equation gives:
\[ r + 2r = 36 + 6 \]
\[ 3r = 42 \]
\[ r = \frac{42}{3} = 14 \]

### Final Answer:
Thus, the value of \( r \) is:
\[ \boxed{14} \]