What is the vlaue of
`1+i^(2)+i^(4)+i^(60)+....+i^(100),` where `i=sqrt(-1)` ?

Options

A. 0
B. 1
C. `-1`
D. None of these

Explanation

To solve the problem, we need to evaluate the expression:

\[ S = 1 + i^2 + i^4 + i^6 + \ldots + i^{100} \]

where \( i = \sqrt{-1} \).

### Step 1: Identify the powers of \( i \)

The powers of \( i \) 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: Determine the values of \( i^2, i^4, i^6, \ldots, i^{100} \)

We will evaluate the even powers of \( i \):
- \( i^2 = -1 \)
- \( i^4 = 1 \)
- \( i^6 = -1 \)
- \( i^8 = 1 \)
- Continuing this pattern, we see:
- For even \( n \), if \( n \equiv 0 \mod 4 \), then \( i^n = 1 \)
- For even \( n \), if \( n \equiv 2 \mod 4 \), then \( i^n = -1 \)

### Step 3: Count the terms in the series

The series goes from \( i^0 \) to \( i^{100} \), which includes all even powers from \( 0 \) to \( 100 \). The even powers are:
- \( 0, 2, 4, 6, \ldots, 100 \)

This is an arithmetic series where:
- First term \( a = 0 \)
- Last term \( l = 100 \)
- Common difference \( d = 2 \)

To find the number of terms \( n \):
\[ n = \frac{l - a}{d} + 1 = \frac{100 - 0}{2} + 1 = 50 + 1 = 51 \]

### Step 4: Calculate the contributions from each type of term

From the sequence of even powers:
- There are \( 26 \) terms where \( n \equiv 0 \mod 4 \) (i.e., \( i^0, i^4, i^8, \ldots, i^{100} \)):
- These contribute \( 1 \) each, so total contribution = \( 26 \times 1 = 26 \).

- There are \( 25 \) terms where \( n \equiv 2 \mod 4 \) (i.e., \( i^2, i^6, i^{10}, \ldots, i^{98} \)):
- These contribute \( -1 \) each, so total contribution = \( 25 \times (-1) = -25 \).

### Step 5: Combine the contributions

Now, we can sum the contributions:
\[ S = 1 + (26 \times 1) + (25 \times -1) = 1 + 26 - 25 = 2 \]

### Step 6: Final result

Thus, the value of the expression is:
\[ \boxed{2} \]