Value of `sum_(k = 1)^(100)(i^(k!) + omega^(k!))`, where i = `sqrt(-1) and omega` is complex cube root of unity , is :

Options

A. `190+omega`
B. `192+omega^(2)`
C. `190+i`
D. `192+i`

Explanation

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

\[ \sum_{k=1}^{100} (i^{k!} + \omega^{k!}) \]

where \(i = \sqrt{-1}\) and \(\omega\) is a complex cube root of unity.

### Step 1: Understanding the components

1. **Complex Unit \(i\)**:
- \(i^1 = i\)
- \(i^2 = -1\)
- \(i^3 = -i\)
- \(i^4 = 1\)
- The powers of \(i\) repeat every 4 terms.

2. **Complex Cube Root of Unity \(\omega\)**:
- The cube roots of unity are \(1, \omega, \omega^2\) where \(\omega = e^{2\pi i / 3}\).
- The properties are:
- \(\omega^3 = 1\)
- \(1 + \omega + \omega^2 = 0\)

### Step 2: Evaluating \(i^{k!}\)

The factorial \(k!\) grows very quickly, and we need to find \(k! \mod 4\) to determine \(i^{k!}\):

- For \(k = 1\): \(1! = 1 \Rightarrow i^{1} = i\)
- For \(k = 2\): \(2! = 2 \Rightarrow i^{2} = -1\)
- For \(k = 3\): \(3! = 6 \Rightarrow i^{6} = (i^4)(i^2) = 1 \cdot (-1) = -1\)
- For \(k \geq 4\): \(k!\) is divisible by 4, so \(i^{k!} = 1\).

Thus, we have:
- For \(k = 1\): \(i^{1!} = i\)
- For \(k = 2\): \(i^{2!} = -1\)
- For \(k = 3\): \(i^{3!} = -1\)
- For \(k = 4\) to \(100\): \(i^{k!} = 1\)

### Step 3: Summing \(i^{k!}\)

Now we sum these values:
\[ \sum_{k=1}^{100} i^{k!} = i + (-1) + (-1) + 1 + 1 + \ldots + 1 \]
- The terms from \(k = 4\) to \(100\) contribute \(97\) terms of \(1\).

Calculating the total:
\[ = i - 1 - 1 + 97 = i + 95 \]

### Step 4: Evaluating \(\omega^{k!}\)

Next, we evaluate \(\sum_{k=1}^{100} \omega^{k!}\):
- For \(k = 1\): \(\omega^{1!} = \omega\)
- For \(k = 2\): \(\omega^{2!} = \omega^2\)
- For \(k = 3\): \(\omega^{3!} = \omega^6 = 1\)
- For \(k \geq 4\): \(k!\) is divisible by \(3\), so \(\omega^{k!} = 1\).

Thus, we have:
\[ \sum_{k=1}^{100} \omega^{k!} = \omega + \omega^2 + 1 + 1 + \ldots + 1 \]
- The terms from \(k = 4\) to \(100\) contribute \(97\) terms of \(1\).

Calculating the total:
\[ = \omega + \omega^2 + 1 + 97 = 0 + 1 + 97 = 98 \]

### Step 5: Final Sum

Now we combine both parts:
\[ \sum_{k=1}^{100} (i^{k!} + \omega^{k!}) = (i + 95) + 98 = i + 193 \]

### Conclusion

The final answer is:
\[ \boxed{193 + i} \]