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_(r=4)^(100)i^(r!)+prod_(r=1)^(101)i^(r)=a+ib, " where " i=sqrt(-1)`, then a+75b, is

Options

A. 11
B. 22
C. 33
D. 44

Answer

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

\[ \sum_{r=4}^{100} i^{r!} + \prod_{r=1}^{101} i^r \]

where \( i = \sqrt{-1} \).

### Step 1: Evaluate the sum \(\sum_{r=4}^{100} i^{r!}\)

The powers of \( i \) cycle every four terms:
- \( i^0 = 1 \)
- \( i^1 = i \)
- \( i^2 = -1 \)
- \( i^3 = -i \)
- \( i^4 = 1 \) (and so on)

To find \( i^{r!} \), we need to determine \( r! \mod 4 \).

#### Step 1.1: Calculate \( r! \mod 4 \)

For \( r \geq 4 \):
- \( 4! = 24 \equiv 0 \mod 4 \)
- \( 5! = 120 \equiv 0 \mod 4 \)
- \( 6! = 720 \equiv 0 \mod 4 \)
- And this continues for all \( r \geq 4 \).

Thus, for all \( r \) from 4 to 100, \( r! \equiv 0 \mod 4 \).

#### Step 1.2: Substitute into the sum

Since \( r! \equiv 0 \mod 4 \), we have:
\[ i^{r!} = i^0 = 1 \]

The sum becomes:
\[ \sum_{r=4}^{100} i^{r!} = \sum_{r=4}^{100} 1 = 100 - 4 + 1 = 97 \]

### Step 2: Evaluate the product \(\prod_{r=1}^{101} i^r\)

Using the property of exponents:
\[ \prod_{r=1}^{101} i^r = i^{\sum_{r=1}^{101} r} \]

#### Step 2.1: Calculate the sum \(\sum_{r=1}^{101} r\)

The sum of the first \( n \) natural numbers is given by:
\[ \sum_{r=1}^{n} r = \frac{n(n+1)}{2} \]
For \( n = 101 \):
\[ \sum_{r=1}^{101} r = \frac{101 \times 102}{2} = 5151 \]

#### Step 2.2: Substitute into the product

Thus, we have:
\[ \prod_{r=1}^{101} i^r = i^{5151} \]

#### Step 2.3: Calculate \( 5151 \mod 4 \)

Now, we need to find \( 5151 \mod 4 \):
\[ 5151 \div 4 = 1287 \quad \text{(remainder 3)} \]
Thus, \( 5151 \equiv 3 \mod 4 \).

Therefore:
\[ i^{5151} = i^3 = -i \]

### Step 3: Combine the results

Now we combine the results from Step 1 and Step 2:
\[ \sum_{r=4}^{100} i^{r!} + \prod_{r=1}^{101} i^r = 97 - i \]

This can be expressed as:
\[ a + ib \quad \text{where } a = 97 \text{ and } b = -1 \]

### Step 4: Calculate \( a + 75b \)

Now we need to find:
\[ a + 75b = 97 + 75(-1) = 97 - 75 = 22 \]

### Final Answer

Thus, the final answer is:
\[ \boxed{22} \]