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Question

Question: \[0.5 - \frac{(0.5)^{2}}{2} + \frac{(0.5)^{3}}{3} - \frac{(0.5)^{4}}{4} + ........\]...

0.5(0.5)22+(0.5)33(0.5)44+........0.5 - \frac{(0.5)^{2}}{2} + \frac{(0.5)^{3}}{3} - \frac{(0.5)^{4}}{4} + ........

A

loge(32)\log_{e}\left( \frac{3}{2} \right)

B

log10(12)\log_{10}\left( \frac{1}{2} \right)

C

loge(n!)\log_{e}(n!)

D

loge(12)\log_{e}\left( \frac{1}{2} \right)

Answer

loge(32)\log_{e}\left( \frac{3}{2} \right)

Explanation

Solution

We know that, xx22+x33x44........=loge(1+x)x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \frac{x^{4}}{4}........\infty = \log_{e}(1 + x)

Putting x = 0.5, we get,

0.5(0.5)22+(0.5)33(0.5)44+........=loge(1+0.5)=loge(32)0.5 - \frac{(0.5)^{2}}{2} + \frac{(0.5)^{3}}{3} - \frac{(0.5)^{4}}{4} + ........\infty = \log_{e}(1 + 0.5) = \log_{e}\left( \frac{3}{2} \right)