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

Question

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

Accepted Answer

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

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

$\begin{gathered} \sum _{n=1}^ \begin{gathered} m=8 \end{gathered}(1-{i}^{n!})=x+iy \ 	herefore \sum _{n=1}^{8}(1-{i}^{n!})=8-(i+{i}^{2}+{i}^{6}+1+1+1+1+1) \ =5-i \ 	herefore x=5,y=-1 \ x+y+m=5-1+8=12 \end{gathered}$

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