• # question_answer If  $x,\,y,z$ are three consecutive positive integers, then $\frac{1}{2}{{\log }_{e}}x+\frac{1}{2}{{\log }_{e}}z+\frac{1}{2xz+1}+\frac{1}{3}{{\left( \frac{1}{2xz+1} \right)}^{3}}+....=$ A) ${{\log }_{e}}x$ B) ${{\log }_{e}}y$ C) ${{\log }_{e}}z$ D) None of these

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$.