Question

If the set $R = {(a, b) : a + 5b = 42, a, b \in \mathbb{N}}$ 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:

• A. 8
• B. 12
• C. 4
• D. 5

Answer 1

The Correct Option is B

\[ a + 5b = 42, \quad a, b \in \mathbb{N} \] \[ a = 42 - 5b, \quad b = 1, \quad a = 37 \] \[ b = 2, \quad a = 32 \] \[ b = 3, \quad a = 27 \] \[ \vdots \] \[ b = 8, \quad a = 2 \] R has "8" elements \( \Rightarrow m = 8 \) \[ \sum_{n=1}^{8} (1 - i^n) = x + iy \] For \( n \geq 4, i^n = 1 \) \[ \Rightarrow (1 - i) + (1 - i^2) + (1 - i^3) \] \[ = 1 - i + 1 - 1 + 1 - i \] \[ = 1 - i + 2 + 1 + 1 - i \] \[ = 5 - i = x + iy \] \[ m + x + y = 8 + 5 - 1 = 12 \] 

Answer 2

From \( a + 5b = 42 \), where \( a, b \in \mathbb{N} \), we have:
\[ a = 42 - 5b. \]
Since \( a > 0 \), we require:\[ 42 - 5b > 0 \implies b < \frac{42}{5} \implies b \leq 8. \]
The possible values of \( (a, b) \) are:
\[ (37, 1), (32, 2), (27, 3), (22, 4), (17, 5), (12, 6), (7, 7), (2, 8). \]
Thus, \( m = 8 \).
The sum is:
\[ \sum_{n=1}^{8}(1 - i^n). \]
For \( n \geq 4 \), \( i^n \) repeats every 4 terms:
\[ i^1 = i, \; i^2 = -1, \; i^3 = -i, \; i^4 = 1. \]
Compute:
\[ \sum_{n=1}^{8}(1 - i^n) = (1 - i) + (1 - (-1)) + (1 - (-i)) + (1 - 1) + (1 - i) + (1 - (-1)) + (1 - (-i)) + (1 - 1). \]
Simplify:
\[ = (1 - i) + 2 + (1 + i) + 0 + (1 - i) + 2 + (1 + i) + 0 = 5 - i + i = 5. \]
Thus:
\[ x + y = 5, \; m = 8, \; m + x + y = 8 + 5 - 1 = 12. \]
Final Answer: 12.