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JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

1 Hard
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Application +4 | -1
If x,y,zx,y,z are three consecutive positive integers, then

12log⁡ex+12log⁡ez+12  x  z+1+13(12  x  z+1)3+…\frac{1}{2}{\log _e}x + \frac{1}{2}{\log _e}z + \frac{1}{{2\;x\;z + 1}} + \frac{1}{3}{\left( {\frac{1}{{2\;x\;z + 1}}} \right)^3} + \ldots is equal to
View answer and explanation
Correct option: B
log⁡ey{\log _e}y
Option B (log⁡ey{\log _e}y) is correct because by using the properties of logarithms and the given condition that x, y, and z are consecutive integers, the expression simplifies to log⁡ey{\log _e}y.

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