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, we need to evaluate the summation:

\[ \sum_{n=0}^{100} i^{n!} \]

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

### Step 1: Understand 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 \( n! \) modulo 4

We need to find \( n! \mod 4 \) for \( n = 0, 1, 2, \ldots, 100 \).

- For \( n = 0 \): \( 0! = 1 \) → \( 1 \mod 4 = 1 \)
- For \( n = 1 \): \( 1! = 1 \) → \( 1 \mod 4 = 1 \)
- For \( n = 2 \): \( 2! = 2 \) → \( 2 \mod 4 = 2 \)
- For \( n = 3 \): \( 3! = 6 \) → \( 6 \mod 4 = 2 \)
- For \( n \geq 4 \): \( n! \) will always be divisible by \( 4 \) (since \( 4! = 24 \) and higher factorials include \( 4 \) as a factor) → \( n! \mod 4 = 0 \)

### Step 3: Calculate \( i^{n!} \) for each \( n \)

Now we can evaluate \( i^{n!} \) for \( n = 0, 1, 2, 3, \ldots, 100 \):

- 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 \( n \geq 4 \): \( i^{n!} = i^0 = 1 \)

### Step 4: Count the contributions to the summation

Now, we can sum these contributions:

- For \( n = 0 \): contributes \( i \)
- For \( n = 1 \): contributes \( i \)
- For \( n = 2 \): contributes \( -1 \)
- For \( n = 3 \): contributes \( -1 \)
- For \( n = 4 \) to \( n = 100 \): contributes \( 1 \) for each of the 97 terms (from \( n = 4 \) to \( n = 100 \))

### Step 5: Calculate the total sum

Now we can sum everything:

\[ \text{Total} = i + i - 1 - 1 + 97 \cdot 1 \]
\[ = 2i - 2 + 97 \]
\[ = 95 + 2i \]

### Final Result

Thus, the value of the summation is:

\[ \boxed{95 + 2i} \]