If the set $R=\{(a, b): a+5 b=42, a, b \in \mathbb{N}\}$ has $m$ elements and $\sum_\limits{n=1}^m\left(1-i^{n !}\right)=x+i y$, where $i=\sqrt{-1}$, then the value of $m+x+y$ is
Solution
<p>$R=\{(a, b): a+5 b=42\}$</p>
<p>Then $R=\{(2,8),(7,7),(12,6),(17,5),(22,4),(27, 3),(32,2),(37,1)\}$</p>
<p>$$\begin{aligned}
& \text { and } \sum_{n=1}^{\substack{m=8}}\left(1-i^{n!}\right)=x+i y \\
& \therefore \sum_{n=1}^8\left(1-i^{n!}\right)=8-\left(i+i^2+i^6+1+1+1+1+1\right) \\
& =5-i \\
& \therefore x=5, y=-1 \\
& x+y+m=5-1+8=12
\end{aligned}$$</p>
About this question
Subject: Mathematics · Chapter: Complex Numbers and Quadratic Equations · Topic: Complex Numbers and Argand Plane
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