If the co-efficient of `x^(2r)` is greater than half of the co-efficient of `x^(2r+1)` in the expansion of `(1+x)^(15)` , then the possible value of 'r' equal to :

Options

A. 5
B. 6
C. 7
D. 8

Answer

To solve the problem, we need to find the values of \( r \) such that the coefficient of \( x^{2r} \) in the expansion of \( (1+x)^{15} \) is greater than half of the coefficient of \( x^{2r+1} \).

### Step-by-Step Solution:

1. **Identify the Coefficients**:
The coefficient of \( x^{2r} \) in \( (1+x)^{15} \) is given by \( \binom{15}{2r} \) and the coefficient of \( x^{2r+1} \) is given by \( \binom{15}{2r+1} \).

2. **Set Up the Inequality**:
We need to set up the inequality:
\[ \binom{15}{2r} > \frac{1}{2} \binom{15}{2r+1} \]

3. **Multiply Both Sides by 2**:
To eliminate the fraction, multiply both sides by 2:
\[ 2 \binom{15}{2r} > \binom{15}{2r+1} \]

4. **Use the Property of Binomial Coefficients**:
We know that:
\[ \binom{n}{k} = \frac{n-k+1}{k} \binom{n}{k-1} \]
Using this property, we can express \( \binom{15}{2r+1} \) in terms of \( \binom{15}{2r} \):
\[ \binom{15}{2r+1} = \frac{15 - 2r}{2r + 1} \binom{15}{2r} \]

5. **Substitute into the Inequality**:
Substitute this back into the inequality:
\[ 2 \binom{15}{2r} > \frac{15 - 2r}{2r + 1} \binom{15}{2r} \]

6. **Cancel \( \binom{15}{2r} \)**:
Assuming \( \binom{15}{2r} \neq 0 \), we can divide both sides by \( \binom{15}{2r} \):
\[ 2 > \frac{15 - 2r}{2r + 1} \]

7. **Cross Multiply**:
Cross multiplying gives:
\[ 2(2r + 1) > 15 - 2r \]

8. **Expand and Rearrange**:
Expanding and rearranging the inequality:
\[ 4r + 2 > 15 - 2r \]
\[ 4r + 2r > 15 - 2 \]
\[ 6r > 13 \]
\[ r > \frac{13}{6} \]

9. **Find Possible Integer Values**:
Since \( r \) must be an integer, the smallest integer greater than \( \frac{13}{6} \approx 2.17 \) is \( 3 \). Thus, \( r \) can take values \( 3, 4, 5, 6, 7, \) and so on.

10. **Check the Upper Limit**:
Since \( 2r \) must be less than or equal to \( 15 \) (as \( (1+x)^{15} \) only goes up to \( x^{15} \)), we have:
\[ 2r \leq 15 \implies r \leq 7.5 \]
Therefore, the maximum integer value for \( r \) is \( 7 \).

### Conclusion:
The possible integer values of \( r \) are \( 3, 4, 5, 6, 7 \).