The value of `sum_(n=0)^(100)i^(n!)` equals (where `i=sqrt(-1))`

Options

A. `-1`
B. `i`
C. `2i+95`
D. `97+i`

Explanation

To solve the problem \( \sum_{n=0}^{100} i^{n!} \), where \( i = \sqrt{-1} \), we will evaluate the powers of \( i \) for factorial values of \( n \).

### Step-by-Step Solution:

1. **Understanding the Powers of \( i \)**:
The powers of \( i \) cycle every 4 terms:
- \( i^0 = 1 \)
- \( i^1 = i \)
- \( i^2 = -1 \)
- \( i^3 = -i \)
- \( i^4 = 1 \) (and the cycle repeats)

2. **Evaluating Factorials**:
We need to evaluate \( i^{n!} \) for \( n = 0 \) to \( 100 \). The factorial \( n! \) grows very quickly:
- \( 0! = 1 \)
- \( 1! = 1 \)
- \( 2! = 2 \)
- \( 3! = 6 \)
- \( 4! = 24 \)
- \( 5! = 120 \)
- For \( n \geq 4 \), \( n! \) will be a multiple of 4.

3. **Calculating Each Term**:
- For \( n = 0 \): \( i^{0!} = i^1 = i \)
- For \( n = 1 \): \( i^{1!} = i^1 = i \)
- For \( n = 2 \): \( i^{2!} = i^2 = -1 \)
- For \( n = 3 \): \( i^{3!} = i^6 = (i^4)(i^2) = 1 \cdot (-1) = -1 \)
- For \( n \geq 4 \): Since \( n! \) is a multiple of 4, \( i^{n!} = i^0 = 1 \).

4. **Summing the Values**:
Now we sum the contributions:
- From \( n = 0 \): \( i \)
- From \( n = 1 \): \( i \)
- From \( n = 2 \): \( -1 \)
- From \( n = 3 \): \( -1 \)
- From \( n = 4 \) to \( n = 100 \): Each contributes \( 1 \).

The total number of terms from \( n = 4 \) to \( n = 100 \) is \( 100 - 4 + 1 = 97 \).

5. **Final Calculation**:
- Contribution from \( n = 0 \) and \( n = 1 \): \( i + i = 2i \)
- Contribution from \( n = 2 \) and \( n = 3 \): \( -1 - 1 = -2 \)
- Contribution from \( n = 4 \) to \( n = 100 \): \( 97 \cdot 1 = 97 \)

Therefore, the total sum is:
\[ 2i - 2 + 97 = 2i + 95 \]

### Final Answer:
The value of \( \sum_{n=0}^{100} i^{n!} \) is \( 2i + 95 \).