If x, y, z are three consecutive positive int | vedclass

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(B) Since $x, y, z$ are three consecutive positive integers,we have $y = x + 1$ and $z = x + 2$,so $z - x = 2$.Also,$2y = x + z$,which implies $y^2 =

Vedclass · Accepted Answer

$\log_e y$

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(B) Since $x, y, z$ are three consecutive positive integers,we have $y = x + 1$ and $z = x + 2$,so $z - x = 2$.Also,$2y = x + z$,which implies $y^2 = xz + 1$ (since $y^2 - xz = (x+1)^2 - x(x+2) = x^2 + 2x + 1 - x^2 - 2x = 1$).Let $S = \frac{1}{2}\log_e x + \frac{1}{2}\log_e z + \sum_{n=1, 3, 5, \dots} \frac{1}{n} \left( \frac{1}{2xz + 1} \right)^n$.Using the logarithmic series expansion $\frac{1}{2} \log_e \left( \frac{1+u}{1-u} \right) = u + \frac{u^3}{3} + \frac{u^5}{5} + \dots$,where $u = \frac{1}{2xz+1}$.$S = \frac{1}{2} \log_e (xz) + \log_e \left( \frac{1 + \frac{1}{2xz+1}}{1 - \frac{1}{2xz+1}} \right) = \frac{1}{2} \log_e (xz) + \log_e \left( \frac{2xz+2}{2xz} \right) = \frac{1}{2} \log_e (xz) + \log_e \left( \frac{xz+1}{xz} \right)$.$S = \frac{1}{2} \log_e (xz) + \log_e (xz+1) - \log_e (xz) = \log_e (xz+1) - \frac{1}{2} \log_e (xz) = \log_e (y^2) - \frac{1}{2} \log_e (xz)$.Since $xz = y^2 - 1$,this simplifies to $\log_e y$.

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 + \dots = $

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