If the coefficients of `(2r+4)^("th")` term and `(3r+4)^("th")` term in the expansion of `(1+x)^(21)` are equal, find r.

Answer

To solve the problem, we need to find the value of \( r \) such that the coefficients of the \( (2r + 4)^{th} \) term and the \( (3r + 4)^{th} \) term in the expansion of \( (1 + x)^{21} \) 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:
\[ T_{k+1} = \binom{n}{k} x^k \]
For \( n = 21 \), the general term becomes:
\[ T_{k+1} = \binom{21}{k} x^k \]

2. **Find the Coefficient of the \( (2r + 4)^{th} \) Term**:
The \( (2r + 4)^{th} \) term corresponds to \( k = 2r + 3 \) (since the term number is \( k + 1 \)):
\[ T_{2r + 4} = \binom{21}{2r + 3} x^{2r + 3} \]
The coefficient is:
\[ \text{Coefficient of } T_{2r + 4} = \binom{21}{2r + 3} \]

3. **Find the Coefficient of the \( (3r + 4)^{th} \) Term**:
The \( (3r + 4)^{th} \) term corresponds to \( k = 3r + 3 \):
\[ T_{3r + 4} = \binom{21}{3r + 3} x^{3r + 3} \]
The coefficient is:
\[ \text{Coefficient of } T_{3r + 4} = \binom{21}{3r + 3} \]

4. **Set the Coefficients Equal**:
According to the problem, we have:
\[ \binom{21}{2r + 3} = \binom{21}{3r + 3} \]

5. **Use the Property of Binomial Coefficients**:
From the property of binomial coefficients, we know that:
\[ \binom{n}{k} = \binom{n}{n-k} \]
This gives us two cases to consider:
- Case 1: \( 2r + 3 = 3r + 3 \)
- Case 2: \( 2r + 3 = 21 - (3r + 3) \)

6. **Solve Case 1**:
\[ 2r + 3 = 3r + 3 \]
Simplifying this, we find:
\[ 2r = 3r \implies r = 0 \]

7. **Solve Case 2**:
\[ 2r + 3 = 21 - (3r + 3) \]
Simplifying this, we get:
\[ 2r + 3 = 21 - 3r - 3 \]
\[ 2r + 3 = 18 - 3r \]
\[ 2r + 3r = 18 - 3 \]
\[ 5r = 15 \implies r = 3 \]

8. **Conclusion**:
The values of \( r \) that satisfy the equation are:
\[ r = 0 \quad \text{or} \quad r = 3 \]