If the set `R = {(a, b) : a + 5b = 42, a, b in N}`has `m` elements and `sum_(n=1)^m(1 – i^(n!)) = x + iy`, where `i = sqrt (-1)`, then the value of `m + x + y` is

Options

A. 5
B. 12
C. 4
D. 8

Answer

To solve the problem, we will follow the steps outlined in the video transcript and provide a detailed solution.

### Step 1: Determine the set R
We start with the equation given in the set \( R = \{(a, b) : a + 5b = 42, a, b \in \mathbb{N}\} \).

Rearranging the equation gives us:
\[ a = 42 - 5b \]
Since \( a \) and \( b \) are natural numbers, \( a \) must be positive. Therefore, we need:
\[ 42 - 5b > 0 \implies 42 > 5b \implies b < \frac{42}{5} = 8.4 \]
Since \( b \) must be a natural number, the possible values for \( b \) are \( 1, 2, 3, 4, 5, 6, 7, 8 \).

### Step 2: Calculate corresponding values of a
Now we will calculate the corresponding values of \( a \) for each value of \( b \):
- For \( b = 1 \): \( a = 42 - 5(1) = 37 \) → \( (37, 1) \)
- For \( b = 2 \): \( a = 42 - 5(2) = 32 \) → \( (32, 2) \)
- For \( b = 3 \): \( a = 42 - 5(3) = 27 \) → \( (27, 3) \)
- For \( b = 4 \): \( a = 42 - 5(4) = 22 \) → \( (22, 4) \)
- For \( b = 5 \): \( a = 42 - 5(5) = 17 \) → \( (17, 5) \)
- For \( b = 6 \): \( a = 42 - 5(6) = 12 \) → \( (12, 6) \)
- For \( b = 7 \): \( a = 42 - 5(7) = 7 \) → \( (7, 7) \)
- For \( b = 8 \): \( a = 42 - 5(8) = 2 \) → \( (2, 8) \)

### Step 3: Count the number of elements in R
The values of \( b \) from \( 1 \) to \( 8 \) give us \( 8 \) pairs in the set \( R \). Thus, \( m = 8 \).

### Step 4: Evaluate the summation
Next, we need to evaluate:
\[ \sum_{n=1}^{m} (1 - i^{{n!}}) \]
Substituting \( m = 8 \):
\[ \sum_{n=1}^{8} (1 - i^{{n!}}) \]

Now we calculate \( i^{n!} \) for \( n = 1, 2, \ldots, 8 \):
- \( n = 1 \): \( 1 - i^{1!} = 1 - i \)
- \( n = 2 \): \( 1 - i^{2!} = 1 - i^2 = 1 + 1 = 2 \)
- \( n = 3 \): \( 1 - i^{3!} = 1 - i^6 = 1 - 1 = 0 \)
- \( n = 4 \): \( 1 - i^{4!} = 1 - i^{24} = 1 - 1 = 0 \)
- \( n = 5 \): \( 1 - i^{5!} = 1 - i^{120} = 1 - 1 = 0 \)
- \( n = 6 \): \( 1 - i^{6!} = 1 - i^{720} = 1 - 1 = 0 \)
- \( n = 7 \): \( 1 - i^{7!} = 1 - i^{5040} = 1 - 1 = 0 \)
- \( n = 8 \): \( 1 - i^{8!} = 1 - i^{40320} = 1 - 1 = 0 \)

### Step 5: Sum the results
Now we sum these results:
\[ (1 - i) + 2 + 0 + 0 + 0 + 0 + 0 + 0 = 3 - i \]
This gives us \( x = 3 \) and \( y = -1 \).

### Step 6: Calculate \( m + x + y \)
Now we calculate:
\[ m + x + y = 8 + 3 - 1 = 10 \]

### Final Answer
Thus, the value of \( m + x + y \) is \( \boxed{10} \).