If the set [R=(a, b): a+5 b=42, a, b N] has m elements and [_n=1^m(1-i^n !)=x+i y], where …

Question

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

Accepted Answer

$$R=\{(a, b): a+5 b=42\}$$

Then $$R=\{(2,8),(7,7),(12,6),(17,5),(22,4),(27, 3),(32,2),(37,1)\}$$

$$\begin{aligned}
& \text { and } \sum_{n=1}^{\substack{m=8}}\left(1-i^{n!}\right)=x+i y \
& \therefore \sum_{n=1}^8\left(1-i^{n!}\right)=8-\left(i+i^2+i^6+1+1+1+1+1\right) \
& =5-i \
& \therefore x=5, y=-1 \
& x+y+m=5-1+8=12
\end{aligned}$$

Solutions · 1