To solve the problem, we need to evaluate the sum:
\[
S = \sum_{k=1}^{100} (i^{k!} + \omega^{k!})
\]
where \(i = \sqrt{-1}\) and \(\omega\) is a complex cube root of unity.
### Step 1: Separate the summation
We can separate the summation into two parts:
\[
S = \sum_{k=1}^{100} i^{k!} + \sum_{k=1}^{100} \omega^{k!}
\]
### Step 2: Evaluate \( \sum_{k=1}^{100} i^{k!} \)
We start with the first summation:
- For \(k=1\), \(i^{1!} = i\)
- For \(k=2\), \(i^{2!} = i^{2} = -1\)
- For \(k=3\), \(i^{3!} = i^{6} = (i^{4})i^{2} = 1 \cdot (-1) = -1\)
- For \(k=4\), \(i^{4!} = i^{24} = (i^{4})^{6} = 1^{6} = 1\)
Notice that \(i^{k!}\) will cycle through \(i, -1, -i, 1\) based on the value of \(k!\) modulo 4.
For \(k \geq 4\), \(k!\) is always a multiple of 4, which means \(i^{k!} = 1\).
Thus, we can summarize:
- \(k = 1: i\)
- \(k = 2: -1\)
- \(k = 3: -1\)
- \(k \geq 4: 1\) (for \(k = 4\) to \(100\))
Counting the contributions:
- From \(k=4\) to \(100\) gives \(97\) terms of \(1\).
- Total contribution from \(k=4\) to \(100\) is \(97 \cdot 1 = 97\).
So the total for the first summation is:
\[
i - 1 - 1 + 97 = i + 95
\]
### Step 3: Evaluate \( \sum_{k=1}^{100} \omega^{k!} \)
Next, we evaluate the second summation:
The complex cube roots of unity satisfy:
\[
\omega^3 = 1 \quad \text{and} \quad 1 + \omega + \omega^2 = 0
\]
- 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 also a multiple of 3, so \(\omega^{k!} = 1\).
Thus, we can summarize:
- \(k = 1: \omega\)
- \(k = 2: \omega^{2}\)
- \(k = 3: 1\)
- \(k \geq 4: 1\) (for \(k = 4\) to \(100\))
Counting the contributions:
- From \(k=4\) to \(100\) gives \(97\) terms of \(1\).
- Total contribution from \(k=4\) to \(100\) is \(97 \cdot 1 = 97\).
So the total for the second summation is:
\[
\omega + \omega^{2} + 1 + 97 = 0 + 97 = 97
\]
### Step 4: Combine the results
Now we combine the results from both summations:
\[
S = (i + 95) + 97 = i + 192
\]
### Final Answer
Thus, the value of the summation is:
\[
\boxed{i + 192}
\]
