Sequence And Series Question 564
Question: 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}{{( \frac{1}{2xz+1} )}^{3}}+….= $
Options:
A) $ {\log_{e}}x $
B) $ {\log_{e}}y $
C) $ {\log_{e}}z $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
Since $ x,y,z $ are three consecutive positive integers, therefore $ 2y=x+z $ .
$ \Rightarrow 4y^{2}={{(x+z)}^{2}}\Rightarrow 4y^{2}={{(x-z)}^{2}}+4xz $
$ \Rightarrow 4y^{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}{{( \frac{1}{1+2xz} )}^{3}}+…. $ $ =\frac{1}{2}[ {\log_{e}}x+{\log_{e}}z . $ $ +2. { ( \frac{1}{1+2xz} )+\frac{1}{3}{{( \frac{1}{1+2xz} )}^{3}}+…. } ] $ $ =\frac{1}{2}[ {\log_{e}}xz+{\log_{e}}( \frac{1+\frac{1}{1+2xz}}{1-\frac{1}{1+2xz}} ) ] $ $ =\frac{1}{2}[ {\log_{e}}xz+{\log_{e}}( \frac{1+xz}{xz} ) ] $ $ =\frac{1}{2}{\log_{e}}(1+xz)=\frac{1}{2}{\log_{e}}y^{2} $ $ ={\log_{e}}y $ .
BETA
