Statement-1: The integeral part of `(8+3sqrt(7))^(20)` is even.
Statement-2: The sum of the last eight coefficients in the expansion of `(1+x)^(16)` is `2^(15)`.
Statement-3: if `R(5sqrt(5)+11)^(2n+1)=[R]+F`, where [R] denotes the greatest integer in R, then `RF=2^(2n+1)`.

Options

A. FFF
B. FFT
C. TFF
D. TFT

Answer

To solve the given problem, we need to evaluate the three statements one by one and determine their validity.

### Statement 1: The integral part of \( (8 + 3\sqrt{7})^{20} \) is even.

1. **Calculate \( (8 + 3\sqrt{7})^{20} + (8 - 3\sqrt{7})^{20} \)**:
- Let \( a = 8 + 3\sqrt{7} \) and \( b = 8 - 3\sqrt{7} \).
- Notice that \( b < 1 \) since \( 3\sqrt{7} \) is approximately \( 7.94 \), making \( 8 - 3\sqrt{7} \approx 0.06 \).
- Therefore, \( b^{20} \) is a very small positive number.

2. **Use the Binomial Theorem**:
- The expression can be expanded using the Binomial Theorem:
\[ a^{20} + b^{20} = \sum_{k=0}^{20} \binom{20}{k} 8^{20-k} (3\sqrt{7})^k + \sum_{k=0}^{20} \binom{20}{k} 8^{20-k} (-3\sqrt{7})^k \]
- This simplifies to:
\[ a^{20} + b^{20} = 2 \sum_{k \text{ even}} \binom{20}{k} 8^{20-k} (3\sqrt{7})^k \]

3. **Determine the integral part**:
- The integral part of \( a^{20} \) is given by:
\[ \lfloor a^{20} \rfloor = a^{20} + b^{20} - b^{20} \approx a^{20} + \text{small number} \]
- Since \( b^{20} \) is very small, it does not affect the integral part significantly.

4. **Conclusion**:
- Since \( a^{20} + b^{20} \) is an even number (as it is twice the sum of even indexed coefficients), the integral part of \( a^{20} \) is odd.
- Thus, **Statement 1 is false**.

### Statement 2: The sum of the last eight coefficients in the expansion of \( (1+x)^{16} \) is \( 2^{15} \).

1. **Identify the coefficients**:
- The last eight coefficients of \( (1+x)^{16} \) are \( \binom{16}{8}, \binom{16}{9}, \ldots, \binom{16}{15}, \binom{16}{16} \).

2. **Use the property of binomial coefficients**:
- The sum of coefficients from \( k=8 \) to \( k=16 \) can be calculated as:
\[ \sum_{k=8}^{16} \binom{16}{k} = 2^{16-1} = 2^{15} \]
- This is because the total sum of coefficients \( (1+x)^{16} \) is \( 2^{16} \) and the coefficients from \( k=0 \) to \( k=7 \) sum to \( 2^{15} \).

3. **Conclusion**:
- Therefore, **Statement 2 is true**.

### Statement 3: If \( R = (5\sqrt{5} + 11)^{2n+1} = [R] + F \), then \( RF = 2^{2n+1} \).

1. **Analyze the expression**:
- Let \( R = (5\sqrt{5} + 11)^{2n+1} \) and \( F \) be the fractional part.
- We also consider \( (5\sqrt{5} - 11)^{2n+1} \), which is a very small positive number.

2. **Evaluate \( R + (5\sqrt{5} - 11)^{2n+1} \)**:
- The sum can be expressed as:
\[ R + (5\sqrt{5} - 11)^{2n+1} = \text{integer} \]
- Thus, \( R + (5\sqrt{5} - 11)^{2n+1} = [R] + F + \text{small number} \).

3. **Determine the product**:
- The product \( RF \) involves the fractional part and the integer part, which does not simplify to \( 2^{2n+1} \) based on the previous calculations.

4. **Conclusion**:
- Therefore, **Statement 3 is false**.

### Final Conclusion:
- **Statement 1**: False
- **Statement 2**: True
- **Statement 3**: False