If the set $R=\{(a, b): a+5 b=42, a, b \in \mathbb{N}\}$ has $m$ elements and $\sum_{n=1}^m\left(1-i^{n…
If the set $R=\{(a, b): a+5 b=42, a, b \in \mathbb{N}\}$ has $m$ elements and $\sum_{n=1}^m\left(1-i^{n !}\right)=x+i y$, where $i=\sqrt{-1}$, then the value of $m+x+y$ is
12
4
8
5
Solution
$\begin{aligned} & a+5 b=42, a, b \in N \\ & a=42-5 b, b=1, a=37 \\ & b=2, a=32 \\ & b=3, a=27 \\ & \vdots \\ & b=8, a=2\end{aligned}$
$\mathrm{R}$ has " 8 " elements $\Rightarrow \mathrm{m}=8$
$\begin{aligned} & \sum_{n=1}^8\left(1-i^{n !}\right)=x+i y \\ & \text { for } n \geq 4, i^{n !}=1 \\ & \Rightarrow(1-i)+\left(1-i^{2 !}\right)+\left(1-i^{3 !}\right) \\ & =1-I+2+1+1 \\ & =5-I=x+i y \\ & m+x+y=8+5-1=12\end{aligned}$