Question

R = (a, b) : a + 5b = 42 and a, b ∈ N has m elements and \(\sum\limits_{n=1}^{m}(1+i^{n!})\) = x + iy (where i = √-1). Find x + y + m.

• A. 20
• B. 12
• C. 8
• D. 13

Answer 1

The Correct Option is A

To solve this problem, we need to break it into two parts: finding the number of elements in the relation \( R \) and evaluating the summation.

Step 1: Identify the Number of Elements in Relation \( R \)

The relation \( R = (a, b) : a + 5b = 42 \) where \( a, b \in \mathbb{N} \) (Natural numbers), implies that for each valid \( b \), there is a corresponding \( a \) such that \( a \) is a natural number.

Rewriting the equation for \( a \) gives: 

\(a = 42 - 5b.\)

Since \( a \) and \( b \) must be natural numbers, \( 42 - 5b > 0 \). Solving gives:

\(42 > 5b \Rightarrow b < \frac{42}{5} = 8.4.\)

Since \( b \) is a natural number, \( b \) can take values from 1 to 8.

Verification with integer numbers:

1. For \( b = 1 \), \( a = 42 - 5 \times 1 = 37 \)
2. For \( b = 2 \), \( a = 42 - 5 \times 2 = 32 \)
3. For \( b = 3 \), \( a = 42 - 5 \times 3 = 27 \)
4. For \( b = 4 \), \( a = 42 - 5 \times 4 = 22 \)
5. For \( b = 5 \), \( a = 42 - 5 \times 5 = 17 \)
6. For \( b = 6 \), \( a = 42 - 5 \times 6 = 12 \)
7. For \( b = 7 \), \( a = 42 - 5 \times 7 = 7 \)
8. For \( b = 8 \), \( a = 42 - 5 \times 8 = 2 \)

All these values satisfy the condition \( a \in \mathbb{N} \). Thus, \( m = 8 \).

Step 2: Evaluate the Summation

We are given:

\(\sum\limits_{n=1}^{m}(1+i^{n!})\) = x +\)

where \( i = \sqrt{-1} \) and \( n! \) is the factorial of \( n \). Since \( i^4 = 1 \), \( i^n \) cycles every 4 terms (i.e., \( i, -1, -i, 1 \)). Calculating for \( n \) from 1 to 8:

• For \( n = 1 \), \( i^{1!} = i \) → \( 1 + i \)
• For \( n = 2 \), \( i^{2!} = i^2 = -1 \) → \( 0 \)
• For \( n = 3 \), \( i^{3!} = i^6 = -1 \) → \( 0 \)
• For \( n = 4 \), \( i^{4!} = i^{24} = 1 \) → \( 2 \)
• For \( n = 5 \), \( i^{5!} = i^{120} = 1 \) → \( 2 \)
• For \( n = 6 \), \( i^{6!} = i^{720} = 1 \) → \( 2 \)
• For \( n = 7 \), \( i^{7!} = i^{5040} = 1 \) → \( 2 \)
• For \( n = 8 \), \( i^{8!} = i^{40320} = 1 \) → \( 2 \)

Summing them:

• Real Part (x): \( 1 + 2 + 2 + 2 + 2 + 2 = 11 \)
• Imaginary Part (y): \( 1 + 0 + 0 + 0 + 0 + 0 = 1 \)

Thus, \( x = 11 \) and \( y = 1 \).

Final Calculation and Conclusion

We need to find \( x + y + m \).

\(x + y + m = 11 + 1 + 8 = 20.\)

Therefore, the answer is 20.