If in the expansion of `(1 + x)^(15)` , the coefficients of `(r -1)^(th)` and `(2r + 3)^(th)` terms are equal, then the value of r is

Options

A. 5
B. 6
C. 4
D. 3

Answer

To solve the problem, we need to find the value of \( r \) such that the coefficients of the \( (r-1)^{th} \) and \( (2r+3)^{th} \) terms in the expansion of \( (1 + x)^{15} \) are equal.

### Step-by-Step Solution:

1. **Identify the General Term**:
The general term in the expansion of \( (1 + x)^{15} \) is given by:
\[ T_k = \binom{15}{k} x^k \]
where \( k \) is the term number starting from \( 0 \).

2. **Coefficients of the Required Terms**:
- The coefficient of the \( (r-1)^{th} \) term is:
\[ \text{Coefficient of } T_{r-1} = \binom{15}{r-1} \]
- The coefficient of the \( (2r+3)^{th} \) term is:
\[ \text{Coefficient of } T_{2r+3} = \binom{15}{2r+3} \]

3. **Set the Coefficients Equal**:
Since the coefficients are equal, we have:
\[ \binom{15}{r-1} = \binom{15}{2r+3} \]

4. **Using the Property of Binomial Coefficients**:
The property states that \( \binom{n}{k} = \binom{n}{n-k} \). Therefore, we can set up two equations:
- \( r - 1 = 2r + 3 \)
- \( r - 1 = 15 - (2r + 3) \)

5. **Solve the First Equation**:
From \( r - 1 = 2r + 3 \):
\[ r - 2r = 3 + 1 \implies -r = 4 \implies r = -4 \]
Since \( r \) cannot be negative, we discard this solution.

6. **Solve the Second Equation**:
From \( r - 1 = 15 - (2r + 3) \):
\[ r - 1 = 15 - 2r - 3 \implies r - 1 = 12 - 2r \]
Rearranging gives:
\[ r + 2r = 12 + 1 \implies 3r = 13 \implies r = 5 \]

7. **Conclusion**:
The value of \( r \) is \( 5 \).