Statement -1 : The expression `((2i)/(1+i))^(n)` is a positive integer for all the values of n.
and
Statement -2 : Here n=8 is the least positive for which the above expression is a positive integer.

Options

A. Statement -1 is True, Statement -2 is True, Statement -2 is a correct explanation for statement -6
B. Statement -1 is True, Statement -2 is True , Statement -2 is NOT a correct explanation for Statement -6
C. Statement -1 is True, Statement -2 is False
D. Statement -1 is Flase, Statement -2 is True

Answer

To solve the problem, we need to analyze the two statements regarding the expression \(\left(\frac{2i}{1+i}\right)^n\).

### Step 1: Simplifying the Expression
We start with the expression:
\[ \frac{2i}{1+i} \]
To simplify this, we can multiply the numerator and the denominator by the conjugate of the denominator:
\[ \frac{2i(1-i)}{(1+i)(1-i)} = \frac{2i - 2i^2}{1^2 - i^2} = \frac{2i + 2}{1 + 1} = \frac{2 + 2i}{2} = 1 + i \]
Thus, we have:
\[ \frac{2i}{1+i} = 1 + i \]

### Step 2: Raising to the Power \(n\)
Now, we need to evaluate:
\[ (1+i)^n \]
To determine if this expression is a positive integer for all \(n\), we can express \(1+i\) in polar form. The modulus of \(1+i\) is:
\[ |1+i| = \sqrt{1^2 + 1^2} = \sqrt{2} \]
The argument (angle) \(\theta\) is:
\[ \theta = \tan^{-1}\left(\frac{1}{1}\right) = \frac{\pi}{4} \]
Thus, we can write:
\[ 1+i = \sqrt{2} \left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right) \]
Using De Moivre's theorem, we can express \((1+i)^n\) as:
\[ (1+i)^n = (\sqrt{2})^n \left(\cos\frac{n\pi}{4} + i\sin\frac{n\pi}{4}\right) = 2^{n/2} \left(\cos\frac{n\pi}{4} + i\sin\frac{n\pi}{4}\right) \]

### Step 3: Analyzing the Result
For \((1+i)^n\) to be a positive integer, the imaginary part must be zero, and the real part must be positive. This occurs when:
\[ \sin\frac{n\pi}{4} = 0 \quad \text{and} \quad \cos\frac{n\pi}{4} > 0 \]
The sine function is zero when:
\[ \frac{n\pi}{4} = k\pi \quad \text{for } k \in \mathbb{Z} \implies n = 4k \]
The cosine function is positive when:
\[ \frac{n\pi}{4} \in (-\frac{\pi}{2}, \frac{\pi}{2}) \quad \text{or } \frac{n}{4} \in (0, 1) \implies n \in (0, 4) \]
Thus, the smallest positive integer \(n\) that satisfies both conditions is \(n=8\).

### Conclusion
- **Statement 1** is false because \((1+i)^n\) is not a positive integer for all \(n\).
- **Statement 2** is true because \(n=8\) is indeed the least positive integer for which \((1+i)^n\) is a positive integer.