If the coefficient of `(3r)^(th)` and `(r + 2)^(th)` terms in the expansion of `(1 + x)^(2n)` are equal then n =

Options

A. 2r
B. 3r
C. 3r + 1
D. 2r + 1

Answer

To solve the problem, we need to find the value of \( n \) given that the coefficients of the \( 3r \)th term and the \( (r + 2) \)th term in the expansion of \( (1 + x)^{2n} \) are equal.

### Step-by-Step Solution:

1. **Identify the General Term**:
The general term in the expansion of \( (1 + x)^{2n} \) is given by:
\[ T_k = \binom{2n}{k} x^k \]
where \( k \) is the term number, starting from 0.

2. **Find the Coefficient of the \( 3r \)th Term**:
The coefficient of the \( 3r \)th term (which corresponds to \( k = 3r \)) is:
\[ \text{Coefficient of } T_{3r} = \binom{2n}{3r} \]

3. **Find the Coefficient of the \( (r + 2) \)th Term**:
The coefficient of the \( (r + 2) \)th term (which corresponds to \( k = r + 2 \)) is:
\[ \text{Coefficient of } T_{r+2} = \binom{2n}{r + 2} \]

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

5. **Use the Property of Binomial Coefficients**:
We can use the property of binomial coefficients which states that:
\[ \binom{n}{k} = \binom{n}{n-k} \]
Therefore, we can write:
\[ \binom{2n}{3r} = \binom{2n}{2n - 3r} \]
Setting this equal to \( \binom{2n}{r + 2} \) gives us:
\[ 3r = 2n - (r + 2) \]

6. **Solve for \( n \)**:
Rearranging the equation:
\[ 3r = 2n - r - 2 \]
\[ 3r + r + 2 = 2n \]
\[ 4r + 2 = 2n \]
Dividing the entire equation by 2:
\[ n = 2r + 1 \]

### Final Result:
Thus, the value of \( n \) is:
\[ n = 2r + 1 \]