The value of `1+ i^(2) + i^(4) + i^(6)+"………"i^(2n)` is

Options

A. Positive
B. Negative
C. Zero
D. Cannot be determined

Answer

To solve the problem of finding the value of the series \(1 + i^2 + i^4 + i^6 + \ldots + i^{2n}\), we can break it down step by step.

### Step 1: Identify the powers of \(i\)
The powers of \(i\) (the imaginary unit) cycle every four terms:
- \(i^0 = 1\)
- \(i^1 = i\)
- \(i^2 = -1\)
- \(i^3 = -i\)
- \(i^4 = 1\) (and the cycle repeats)

### Step 2: Analyze the series
The series given is:
\[ 1 + i^2 + i^4 + i^6 + \ldots + i^{2n} \]
This series consists of terms where the exponent is even. Therefore, we can rewrite the powers of \(i\) in the series:
- \(i^2 = -1\)
- \(i^4 = 1\)
- \(i^6 = -1\)
- \(i^8 = 1\)
- and so on.

### Step 3: Determine the pattern
From the pattern, we can see:
- For even powers \(i^{2k}\) where \(k\) is even, \(i^{2k} = 1\).
- For even powers \(i^{2k}\) where \(k\) is odd, \(i^{2k} = -1\).

### Step 4: Count the terms
The number of terms in the series from \(i^0\) to \(i^{2n}\) is \(n + 1\) (since we start counting from \(0\)).

### Step 5: Evaluate the series based on \(n\)
1. **If \(n\) is even**:
- The last term \(i^{2n} = 1\).
- The series will have \(\frac{n}{2} + 1\) terms equal to \(1\) and \(\frac{n}{2}\) terms equal to \(-1\).
- The sum will be:
\[ \left(\frac{n}{2} + 1\right) \cdot 1 + \left(\frac{n}{2}\right) \cdot (-1) = \frac{n}{2} + 1 - \frac{n}{2} = 1 \]

2. **If \(n\) is odd**:
- The last term \(i^{2n} = -1\).
- The series will have \(\frac{n + 1}{2}\) terms equal to \(1\) and \(\frac{n - 1}{2}\) terms equal to \(-1\).
- The sum will be:
\[ \left(\frac{n + 1}{2}\right) \cdot 1 + \left(\frac{n - 1}{2}\right) \cdot (-1) = \frac{n + 1}{2} - \frac{n - 1}{2} = 1 \]
- However, the last term being \(-1\) means the total sum will actually be \(0\).

### Conclusion
Thus, the value of the series depends on whether \(n\) is even or odd:
- If \(n\) is even, the sum is \(1\).
- If \(n\) is odd, the sum is \(0\).

### Final Answer
The value of \(1 + i^2 + i^4 + i^6 + \ldots + i^{2n}\) is:
- \(1\) if \(n\) is even,
- \(0\) if \(n\) is odd.