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:

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

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

### Step 1: Split the Summation
We can split the summation into two parts:

\[ \Sigma_{k=1}^{99} i^{k!} + \Sigma_{k=1}^{99} \omega^{k!} \]

Let’s denote:

\[ A = \Sigma_{k=1}^{99} i^{k!} \quad \text{and} \quad B = \Sigma_{k=1}^{99} \omega^{k!} \]

### Step 2: Calculate \( A = \Sigma_{k=1}^{99} i^{k!} \)
Now, we will calculate \( A \):

- For \( k = 1 \): \( i^{1!} = i^1 = i \)
- For \( k = 2 \): \( i^{2!} = i^2 = -1 \)
- For \( k = 3 \): \( i^{3!} = i^6 = (i^4)(i^2) = 1 \cdot (-1) = -1 \)
- For \( k = 4 \): \( i^{4!} = i^{24} = (i^4)^6 = 1^6 = 1 \)

For \( k \geq 4 \), \( k! \) is a multiple of 4, thus:

\[ i^{k!} = (i^4)^{m} = 1 \quad \text{for some integer } m \]

So, for \( k \geq 4 \), \( i^{k!} = 1 \).

Now, we can summarize the contributions:

- \( A = i + (-1) + (-1) + 1 + 1 + 1 + \ldots + 1 \)
- The number of terms from \( k = 4 \) to \( k = 99 \) is \( 99 - 4 + 1 = 96 \).

Thus:

\[ A = i - 1 - 1 + 1 + 96 = i + 94 \]

### Step 3: Calculate \( B = \Sigma_{k=1}^{99} \omega^{k!} \)
Now, we will calculate \( B \):

Recall that \( \omega \) is a non-real cube root of unity, satisfying \( \omega^3 = 1 \) and \( 1 + \omega + \omega^2 = 0 \).

- For \( k = 1 \): \( \omega^{1!} = \omega \)
- For \( k = 2 \): \( \omega^{2!} = \omega^2 \)
- For \( k = 3 \): \( \omega^{3!} = \omega^6 = (\omega^3)^2 = 1 \)
- For \( k \geq 3 \), \( k! \) is a multiple of 3, thus \( \omega^{k!} = 1 \).

So, we summarize the contributions:

\[ B = \omega + \omega^2 + 1 + 1 + \ldots + 1 \]

The number of terms from \( k = 3 \) to \( k = 99 \) is \( 99 - 3 + 1 = 97 \).

Thus:

\[ B = \omega + \omega^2 + 1 + 96 = 0 + 96 = 96 \]

### Step 4: Combine \( A \) and \( B \)
Now, we can combine \( A \) and \( B \):

\[ \Sigma_{k=1}^{99} (i^{k!} + \omega^{k!}) = A + B = (i + 94) + 96 = i + 190 \]

### Final Answer
The final value of the summation is:

\[ \boxed{190 + i} \]