If the coefficient of `(2r + 4)^(th)` term and `(r - 2)^(th)` term in the expansion of `(1 + x)^18` are equal then find r.

Options

A. 9
B. 4
C. 6
D. 3

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 \( (r - 2)^{th} \) term in the expansion of \( (1 + x)^{18} \) are equal.

### Step-by-Step Solution:

1. **Identify the General Term**:
The \( (n + 1)^{th} \) term in the expansion of \( (1 + x)^n \) is given by:
\[ T_{k+1} = \binom{n}{k} x^k \]
For our case, \( n = 18 \), so the \( (r + 1)^{th} \) term is:
\[ T_{r + 1} = \binom{18}{r} x^r \]

2. **Find the Coefficient of the \( (2r + 4)^{th} \) Term**:
The \( (2r + 4)^{th} \) term corresponds to \( k = 2r + 4 \):
\[ T_{2r + 5} = \binom{18}{2r + 4} x^{2r + 4} \]
Thus, the coefficient is:
\[ \text{Coefficient of } T_{2r + 5} = \binom{18}{2r + 4} \]

3. **Find the Coefficient of the \( (r - 2)^{th} \) Term**:
The \( (r - 2)^{th} \) term corresponds to \( k = r - 2 \):
\[ T_{r - 1} = \binom{18}{r - 2} x^{r - 2} \]
Thus, the coefficient is:
\[ \text{Coefficient of } T_{r - 1} = \binom{18}{r - 2} \]

4. **Set the Coefficients Equal**:
According to the problem, we have:
\[ \binom{18}{2r + 4} = \binom{18}{r - 2} \]

5. **Use the Property of Binomial Coefficients**:
The property \( \binom{n}{k} = \binom{n}{n-k} \) gives us:
\[ 2r + 4 + (r - 2) = 18 \]
Simplifying this:
\[ 3r + 2 = 18 \]
\[ 3r = 16 \]
\[ r = \frac{16}{3} \quad \text{(not an integer, discard)} \]

6. **Consider the Other Case**:
The other case is:
\[ 2r + 4 = 18 - (r - 2) \]
Simplifying this:
\[ 2r + 4 = 20 - r \]
\[ 3r = 16 \]
\[ r = 6 \]

7. **Conclusion**:
The value of \( r \) that satisfies the condition is:
\[ \boxed{6} \]