If `i^(2)-1 and Sigma_(r=1)^(n)(i)^(r ) AA n in N`, is a non - zero real number, then n can be

Options

A. 100
B. 201
C. 302
D. 403

Answer

To solve the problem, we need to analyze the given expression and the conditions under which the sum is a non-zero real number.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We are given that \( i^2 = -1 \) and we need to evaluate the sum \( S = \sum_{r=1}^{n} i^r \) for \( n \in \mathbb{N} \). We want to find the values of \( n \) such that \( S \) is a non-zero real number.

2. **Calculating the Sum**:
The sum \( S \) can be expressed as:
\[ S = i + i^2 + i^3 + \ldots + i^n \]
Using the formula for the sum of a geometric series, we can rewrite this as:
\[ S = \frac{i(1 - i^n)}{1 - i} \]

3. **Simplifying the Denominator**:
We simplify the denominator:
\[ 1 - i = 1 + i^2 = 1 - (-1) = 2 \]
Thus, we have:
\[ S = \frac{i(1 - i^n)}{2} \]

4. **Finding \( i^n \)**:
The powers of \( i \) cycle every 4 terms:
- \( i^1 = i \)
- \( i^2 = -1 \)
- \( i^3 = -i \)
- \( i^4 = 1 \)
- \( i^5 = i \) (and so on)

Therefore, we can express \( i^n \) based on \( n \mod 4 \):
- If \( n \equiv 0 \mod 4 \), \( i^n = 1 \)
- If \( n \equiv 1 \mod 4 \), \( i^n = i \)
- If \( n \equiv 2 \mod 4 \), \( i^n = -1 \)
- If \( n \equiv 3 \mod 4 \), \( i^n = -i \)

5. **Substituting \( i^n \) into \( S \)**:
We substitute \( i^n \) into the expression for \( S \):
- For \( n \equiv 0 \mod 4 \):
\[ S = \frac{i(1 - 1)}{2} = 0 \quad (\text{not valid}) \]
- For \( n \equiv 1 \mod 4 \):
\[ S = \frac{i(1 - i)}{2} = \frac{i - i^2}{2} = \frac{i + 1}{2} \quad (\text{complex}) \]
- For \( n \equiv 2 \mod 4 \):
\[ S = \frac{i(1 + 1)}{2} = \frac{2i}{2} = i \quad (\text{complex}) \]
- For \( n \equiv 3 \mod 4 \):
\[ S = \frac{i(1 + i)}{2} = \frac{i + i^2}{2} = \frac{i - 1}{2} \quad (\text{complex}) \]

6. **Conclusion**:
The only case where \( S \) is a non-zero real number occurs when \( n \equiv 2 \mod 4 \). This means \( n \) can take values like \( 2, 6, 10, 14, \ldots \) (i.e., \( n = 4k + 2 \) for \( k \in \mathbb{N} \)).

### Final Answer:
Thus, \( n \) can be of the form \( 4k + 2 \) where \( k \) is a non-negative integer.