, then
A) \[xyz=x+y+z\]
B) \[xz+yz=xy+z\]
C) \[xy+yz=xz+y\]
D) \[xy+xz=yz+x\]
Correct Answer: B
Solution :
We have \[x=\sum\limits_{n=0}^{\infty }{{{a}^{n}}}=\frac{1}{1-a}\Rightarrow a=\frac{x-1}{x}\] \[y=\sum\limits_{n=0}^{\infty }{{{b}^{n}}}=\frac{1}{1-b}\]\[\Rightarrow \]\[b=\frac{y-1}{y}\] \[z=\sum\limits_{n=0}^{\infty }{{{a}^{n}}{{b}^{n}}=\frac{1}{1-ab}\Rightarrow ab=\frac{z-1}{z}}\] \[\therefore \]\[\frac{x-1}{x}.\frac{y-1}{y}=\frac{z-1}{z}\]\[\Rightarrow \]\[xy+z=zx+yz\].
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