If 0

Question

(A) Given $x(y^3 - 1) = 1$,we have $x = \frac{1}{y^3 - 1}$.Let $k = \frac{1}{x} = y^3 - 1$. Since $0 < y < 2^{1/3}$,we have $0 < y^3 < 2$,which implie

Vedclass · Accepted Answer

$\log \left( \frac{y^3}{2 - y^3} \right)$

Vedclass · Answer

(A) Given $x(y^3 - 1) = 1$,we have $x = \frac{1}{y^3 - 1}$.Let $k = \frac{1}{x} = y^3 - 1$. Since $0 The given series is $S = 2 \left( \frac{1}{x} + \frac{1}{3x^3} + \frac{1}{5x^5} + \dots \right) = 2 \left( k + \frac{k^3}{3} + \frac{k^5}{5} + \dots \right)$.Using the logarithmic expansion $\log \left( \frac{1+k}{1-k} \right) = 2 \left( k + \frac{k^3}{3} + \frac{k^5}{5} + \dots \right)$,we get $S = \log \left( \frac{1+k}{1-k} \right)$.Substituting $k = y^3 - 1$,we get $S = \log \left( \frac{1 + (y^3 - 1)}{1 - (y^3 - 1)} \right) = \log \left( \frac{y^3}{2 - y^3} \right)$.

If $0 < y < 2^{1/3}$ and $x(y^3 - 1) = 1$,then $\frac{2}{x} + \frac{2}{3x^3} + \frac{2}{5x^5} + \dots$ is equal to:

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