Sum of four consecutive powers of i(iota) is zero.
i.e.,`i^(n)+i^(n+1)+i^(n+2)+i^(n+3)=0,forall n in I.`
If `sum_(n=1)^(25)i^(n!)=a+ib, " where " i=sqrt(-1)`, then a-b, is

Options

A. prime number
B. even number
C. composite number
D. perfect number

Answer

To solve the problem, we need to evaluate the sum \( \sum_{n=1}^{25} i^{n!} \) and express it in the form \( a + ib \), where \( i = \sqrt{-1} \). Then, we will find \( a - b \).

### Step-by-Step Solution:

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

This means that \( i^{n} \) can be determined by \( n \mod 4 \).

2. **Finding \( n! \mod 4 \)**:
We need to find \( n! \mod 4 \) for \( n = 1, 2, \ldots, 25 \):
- For \( n = 1 \): \( 1! = 1 \mod 4 \) → \( i^{1!} = i \)
- For \( n = 2 \): \( 2! = 2 \mod 4 \) → \( i^{2!} = -1 \)
- For \( n = 3 \): \( 3! = 6 \mod 4 \) → \( i^{3!} = -i \)
- For \( n = 4 \): \( 4! = 24 \mod 4 \) → \( i^{4!} = 1 \)
- For \( n \geq 4 \): \( n! \) is always divisible by 4, so \( n! \mod 4 = 0 \) → \( i^{n!} = 1 \)

3. **Counting contributions from each \( n \)**:
- For \( n = 1 \): contributes \( i \)
- For \( n = 2 \): contributes \( -1 \)
- For \( n = 3 \): contributes \( -i \)
- For \( n = 4 \): contributes \( 1 \)
- For \( n = 5 \) to \( n = 25 \) (21 terms): each contributes \( 1 \)

4. **Calculating the total sum**:
\[ \text{Total contribution} = i + (-1) + (-i) + 1 + 21 \cdot 1 \]
Simplifying this:
- The imaginary parts: \( i - i = 0 \)
- The real parts: \( -1 + 1 + 21 = 21 \)

Thus, the total sum is:
\[ 21 + 0i \]

5. **Identifying \( a \) and \( b \)**:
From the sum \( 21 + 0i \), we have:
- \( a = 21 \)
- \( b = 0 \)

6. **Finding \( a - b \)**:
\[ a - b = 21 - 0 = 21 \]

### Final Answer:
The value of \( a - b \) is \( \boxed{21} \).