A) \[{{\log }_{e}}x\]
B) \[{{\log }_{e}}y\]
C) \[{{\log }_{e}}z\]
D) None of these
Correct Answer: B
Solution :
Since \[x,y,z\]are three consecutive positive integers, therefore \[2y=x+z\]. \[\Rightarrow \,\,4{{y}^{2}}={{(x+z)}^{2}}\Rightarrow 4{{y}^{2}}={{(x-z)}^{2}}+4xz\] \[\Rightarrow \,\,4{{y}^{2}}={{(-2)}^{2}}+4xz,\,\,\,(\because z-x=-2)\] \[\Rightarrow \,\,{{y}^{2}}=1+xz\] ....(i) Now \[\frac{1}{2}{{\log }_{e}}x+\frac{1}{2}{{\log }_{e}}z+\frac{1}{1+2xz}+\frac{1}{3}{{\left( \frac{1}{1+2xz} \right)}^{3}}+....\] \[=\frac{1}{2}\left[ {{\log }_{e}}x+{{\log }_{e}}z \right.\]\[+2\left. \left\{ \left( \frac{1}{1+2xz} \right)+\frac{1}{3}{{\left( \frac{1}{1+2xz} \right)}^{3}}+.... \right\} \right]\] \[=\frac{1}{2}\left[ {{\log }_{e}}xz+{{\log }_{e}}\left( \frac{1+\frac{1}{1+2xz}}{1-\frac{1}{1+2xz}} \right) \right]\] \[=\frac{1}{2}\left[ {{\log }_{e}}xz+{{\log }_{e}}\left( \frac{1+xz}{xz} \right) \right]\] \[=\frac{1}{2}{{\log }_{e}}(1+xz)=\frac{1}{2}{{\log }_{e}}{{y}^{2}}\]\[={{\log }_{e}}y\].
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