The expression log_{e} 2 + log_{e} left( 1 + | vedclass

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(B) The given series is $S = \log_{e} 2 + \log_{e} \left( \frac{3}{2} \right) + \log_{e} \left( \frac{4}{3} \right) + \dots + \log_{e} \left( \frac{n}

Vedclass · Accepted Answer

$\log_{e} n$

Vedclass · Answer

(B) The given series is $S = \log_{e} 2 + \log_{e} \left( \frac{3}{2} ight) + \log_{e} \left( \frac{4}{3} ight) + \dots + \log_{e} \left( \frac{n}{n - 1} ight)$.Using the property $\log a + \log b = \log(ab)$,we get:$S = \log_{e} \left( 2 	imes \frac{3}{2} 	imes \frac{4}{3} 	imes \dots 	imes \frac{n}{n - 1} ight)$.All intermediate terms cancel out:$S = \log_{e} (n)$.Thus,the correct option is $B$.

The expression $\log_{e} 2 + \log_{e} \left( 1 + \frac{1}{2} \right) + \log_{e} \left( 1 + \frac{1}{3} \right) + \dots + \log_{e} \left( 1 + \frac{1}{n - 1} \right)$ is equal to

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