frac{4}{1 times 3} - frac{6}{2 times 4} + fra | vedclass

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(B) The given series is $S = \frac{4}{1 	imes 3} - \frac{6}{2 	imes 4} + \frac{12}{5 	imes 7} - \frac{14}{6 	imes 8} + \dots$We can rewrite the te

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$\log_e 2$

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(B) The given series is $S = \frac{4}{1 	imes 3} - \frac{6}{2 	imes 4} + \frac{12}{5 	imes 7} - \frac{14}{6 	imes 8} + \dots$We can rewrite the terms as:$S = \left( \frac{1+3}{1 	imes 3} ight) - \left( \frac{2+4}{2 	imes 4} ight) + \left( \frac{5+7}{5 	imes 7} ight) - \left( \frac{6+8}{6 	imes 8} ight) + \dots$$S = \left( \frac{1}{3} + \frac{1}{1} ight) - \left( \frac{1}{4} + \frac{1}{2} ight) + \left( \frac{1}{7} + \frac{1}{5} ight) - \left( \frac{1}{8} + \frac{1}{6} ight) + \dots$Rearranging the terms,we get:$S = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \dots$This is the standard expansion for $\log_e(1+x)$ at $x=1$,which is $\log_e 2$.

$\frac{4}{1 \times 3} - \frac{6}{2 \times 4} + \frac{12}{5 \times 7} - \frac{14}{6 \times 8} + \dots \infty = $

Similar Questions

If $4\left[ {{x^2} + \frac{{{x^6}}}{3} + \frac{{{x^{10}}}}{5} + \dots} \right] = {y^2} + \frac{{{y^4}}}{2} + \frac{{{y^6}}}{3} + \dots$,then

$\frac{2}{1} \cdot \frac{1}{3} + \frac{3}{2} \cdot \frac{1}{9} + \frac{4}{3} \cdot \frac{1}{27} + \frac{5}{4} \cdot \frac{1}{81} + \dots \infty = $

The sum of $\frac{1}{2} + \frac{1}{3} \cdot \frac{1}{2^3} + \frac{1}{5} \cdot \frac{1}{2^5} + \dots \infty$ is

$\frac{m - n}{m + n} + \frac{1}{3}\left( \frac{m - n}{m + n} \right)^3 + \frac{1}{5}\left( \frac{m - n}{m + n} \right)^5 + \dots \infty = $

$\frac{1}{2}x^2 + \frac{2}{3}x^3 + \frac{3}{4}x^4 + \dots \infty = $

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