The value of `i^2 + i^4 + i^6 + i^8....`upto (2n+1) terms , where `i^2` = -1, is equal to:

Options

A. `-1`
B. 1
C. `-i`
D. i

Explanation

To find the value of \( i^2 + i^4 + i^6 + i^8 + \ldots \) up to \( (2n + 1) \) terms, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Series**:
The series given is \( i^2 + i^4 + i^6 + i^8 + \ldots \) up to \( (2n + 1) \) terms.

2. **Recognize the Pattern**:
We can express the terms in the series as:
\[ i^2, i^4, i^6, \ldots, i^{2(2n+1)} \]
This can be rewritten as:
\[ i^2, (i^2)^2, (i^2)^3, \ldots, (i^2)^{(n+1)} \]
Here, the first term \( a = i^2 \) and the common ratio \( r = i^2 \).

3. **Determine the Number of Terms**:
The number of terms in the series is \( n + 1 \) since we start counting from \( i^2 \) to \( i^{2(2n+1)} \).

4. **Use the Formula for the Sum of a Geometric Series**:
The sum \( S_n \) of the first \( n \) terms of a geometric series can be calculated using the formula:
\[ S_n = a \frac{1 - r^n}{1 - r} \]
In our case:
- \( a = i^2 \)
- \( r = i^2 \)
- Number of terms \( n + 1 \)

Thus, we can write:
\[ S = i^2 \frac{1 - (i^2)^{n+1}}{1 - i^2} \]

5. **Substituting Values**:
We know \( i^2 = -1 \), so we substitute:
\[ S = (-1) \frac{1 - (-1)^{n+1}}{1 - (-1)} \]

6. **Simplifying the Denominator**:
The denominator becomes:
\[ 1 - (-1) = 2 \]

7. **Final Expression**:
Now substituting back, we get:
\[ S = (-1) \frac{1 - (-1)^{n+1}}{2} \]

8. **Evaluating the Expression**:
- If \( n + 1 \) is even (i.e., \( n \) is odd), then \( (-1)^{n+1} = 1 \):
\[ S = (-1) \frac{1 - 1}{2} = 0 \]
- If \( n + 1 \) is odd (i.e., \( n \) is even), then \( (-1)^{n+1} = -1 \):
\[ S = (-1) \frac{1 - (-1)}{2} = (-1) \frac{2}{2} = -1 \]

### Conclusion:
Thus, the value of \( i^2 + i^4 + i^6 + \ldots \) up to \( (2n + 1) \) terms is:
- \( 0 \) if \( n \) is odd,
- \( -1 \) if \( n \) is even.