Statement-1: The middle term of `(x+(1)/(x))^(2n)` can exceed `((2n)^(n))/(n!)` for some value of x.
Statement-2: The coefficient of `x^(n)` in the expansion of `(1-2x+3x^(2)-4x^(3)+ . . .)^(-n)` is `(1*3*5 . . .(2n-1))/(n!)*2^(n)`.
Statement-3: The coefficient of `x^(5)` in `(1+2x+3x^(2)+ . . .)^(-3//2)` is 2.1.

Options

A. TTF
B. FTT
C. FTF
D. FFF

Answer

To solve the problem, we will analyze each statement one by one and determine whether they are true or false based on mathematical principles and the Binomial Theorem.

### Statement 1:
**Claim:** The middle term of \( (x + \frac{1}{x})^{2n} \) can exceed \( \frac{(2n)^n}{n!} \) for some value of \( x \).

**Solution:**
1. **Identify the middle term:** For the expansion of \( (x + \frac{1}{x})^{2n} \), the middle term occurs when \( r = n \) (since there are \( 2n + 1 \) terms).
2. **General term formula:** The \( r \)-th term in the expansion is given by:
\[ T_r = \binom{2n}{r} x^{2n-r} \left(\frac{1}{x}\right)^r = \binom{2n}{r} x^{2n - 2r} \]
3. **Middle term for \( r = n \):**
\[ T_n = \binom{2n}{n} x^{2n - 2n} = \binom{2n}{n} \]
4. **Evaluate \( \binom{2n}{n} \):** The value of \( \binom{2n}{n} \) can be approximated using Stirling's approximation:
\[ \binom{2n}{n} \approx \frac{4^n}{\sqrt{\pi n}} \]
5. **Comparison:** We need to check if \( \binom{2n}{n} \) can exceed \( \frac{(2n)^n}{n!} \). Using Stirling's approximation for \( n! \):
\[ n! \approx \sqrt{2 \pi n} \left(\frac{n}{e}\right)^n \]
Therefore, we can show that:
\[ \frac{(2n)^n}{n!} \approx \frac{(2n)^n e^n}{\sqrt{2 \pi n} n^n} = \frac{2^n e^n}{\sqrt{2 \pi n}} \]
As \( n \) grows, \( \binom{2n}{n} \) can exceed \( \frac{(2n)^n}{n!} \).

**Conclusion:** The statement is **True**.

### Statement 2:
**Claim:** The coefficient of \( x^n \) in the expansion of \( (1 - 2x + 3x^2 - 4x^3 + \ldots)^{-n} \) is \( \frac{1 \cdot 3 \cdot 5 \ldots (2n - 1)}{n!} \cdot 2^n \).

**Solution:**
1. **Recognize the series:** The series \( 1 - 2x + 3x^2 - 4x^3 + \ldots \) can be expressed as a power series.
2. **Transform the series:** This series can be rewritten as:
\[ \sum_{k=0}^{\infty} (-1)^k (k+1) x^k = \frac{1}{(1 + 2x)^2} \]
3. **Expansion:** We need to find the coefficient of \( x^n \) in \( \left(\frac{1}{(1 + 2x)^2}\right)^{-n} \).
4. **Using the Binomial Theorem:** The coefficient of \( x^n \) in \( (1 + 2x)^{2n} \) is given by:
\[ \binom{2n}{n} (2)^n \]
5. **Final coefficient:** The coefficient simplifies to:
\[ \frac{(2n)!}{(n!)^2} \cdot 2^n \]
This does not match the proposed coefficient.

**Conclusion:** The statement is **False**.

### Statement 3:
**Claim:** The coefficient of \( x^5 \) in \( (1 + 2x + 3x^2 + \ldots)^{-3/2} \) is 2.1.

**Solution:**
1. **Recognize the series:** The series \( 1 + 2x + 3x^2 + \ldots \) can be expressed as:
\[ \frac{1}{(1 - x)^2} \]
2. **Transform the series:** Thus, \( (1 + 2x + 3x^2 + \ldots)^{-3/2} = \left(\frac{1}{(1-x)^2}\right)^{-3/2} = (1-x)^3 \).
3. **Find the coefficient of \( x^5 \):** The expansion of \( (1-x)^3 \) gives:
\[ \sum_{k=0}^{\infty} \binom{3}{k} (-1)^k x^k \]
The coefficient of \( x^5 \) in this expansion is zero because the polynomial degree is less than 5.

**Conclusion:** The statement is **False**.

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