The value of `Sigma_(k=1)^(99)(i^(k!)+omega^(k!))` is (where, `i=sqrt(-1)` amd `omega` is non - real cube root of unity)

Options

A. `190+omega`
B. `192+omega^(2)`
C. `190+i`
D. `102+i`

Answer

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

\[ S = \sum_{k=1}^{99} (i^{k!} + \omega^{k!}) \]

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

### Step 1: Understanding the Powers of \(i\)

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

The factorial \(k!\) for \(k \geq 4\) will always be a multiple of 4. Thus, for \(k \geq 4\), \(i^{k!} = 1\).

### Step 2: Evaluating \(i^{k!}\) for \(k = 1, 2, 3\)

- For \(k = 1\): \(i^{1!} = i^{1} = i\)
- For \(k = 2\): \(i^{2!} = i^{2} = -1\)
- For \(k = 3\): \(i^{3!} = i^{6} = (i^4)(i^2) = 1 \cdot (-1) = -1\)

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

Now, we can sum these contributions:
- For \(k = 1\): \(i\)
- For \(k = 2\): \(-1\)
- For \(k = 3\): \(-1\)
- For \(k = 4\) to \(99\): \(1\) (96 terms)

So, the total contribution from \(i^{k!}\) is:
\[ S_i = i - 1 - 1 + 96 \cdot 1 = i - 2 + 96 = i + 94 \]

### Step 4: Understanding the Powers of \(\omega\)

The powers of \(\omega\) cycle every 3 terms:
- \(\omega^1 = \omega\)
- \(\omega^2 = \omega^2\)
- \(\omega^3 = 1\)
- \(\omega^4 = \omega\) (and so on)

The factorial \(k!\) for \(k \geq 3\) will always be a multiple of 3. Thus, for \(k \geq 3\), \(\omega^{k!} = 1\).

### Step 5: Evaluating \(\omega^{k!}\) for \(k = 1, 2, 3\)

- For \(k = 1\): \(\omega^{1!} = \omega^{1} = \omega\)
- For \(k = 2\): \(\omega^{2!} = \omega^{2} = \omega^2\)
- For \(k = 3\): \(\omega^{3!} = \omega^{6} = 1\)

### Step 6: Summing \(\omega^{k!}\)

Now, we can sum these contributions:
- For \(k = 1\): \(\omega\)
- For \(k = 2\): \(\omega^2\)
- For \(k = 3\): \(1\)
- For \(k = 4\) to \(99\): \(1\) (96 terms)

So, the total contribution from \(\omega^{k!}\) is:
\[ S_\omega = \omega + \omega^2 + 1 + 96 \cdot 1 = \omega + \omega^2 + 1 + 96 \]
Since \(\omega + \omega^2 + 1 = 0\) (property of cube roots of unity), we have:
\[ S_\omega = 0 + 96 = 96 \]

### Step 7: Final Summation

Now, we combine the contributions from \(i^{k!}\) and \(\omega^{k!}\):
\[ S = S_i + S_\omega = (i + 94) + 96 = i + 190 \]

Thus, the final answer is:
\[ \boxed{190 + i} \]