(0.5) - frac{(0.5)^2}{2} + frac{(0.5)^3}{3} - | vedclass

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Question

(A) The given series is of the form $x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots$ where $x = 0.5$.We know that the expansion of $\log_e(

Vedclass · Accepted Answer

$\log_e \frac{3}{2}$

Vedclass · Answer

(A) The given series is of the form $x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots$ where $x = 0.5$.We know that the expansion of $\log_e(1 + x)$ is given by $\log_e(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots$ for $-1 Substituting $x = 0.5$ into the series,we get:$0.5 - \frac{(0.5)^2}{2} + \frac{(0.5)^3}{3} - \frac{(0.5)^4}{4} + \dots = \log_e(1 + 0.5)$.$= \log_e(1.5) = \log_e(\frac{3}{2})$.

$(0.5) - \frac{(0.5)^2}{2} + \frac{(0.5)^3}{3} - \frac{(0.5)^4}{4} + \dots$

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