Let $$x = -\frac{1}{1!}\cdot\frac{3}{4} + \frac{1}{2!}\cdot(\frac{3}{4})^2 -\frac{1}{3!}\cdot(\frac{3}{4})^3 +$$.... and $$y = x - \frac{x^2}{2} + \frac{x^3}{3} - $$..... then the value of y is
$$\log (1- x)=$$
$$-x -\frac{1}{2}x^2 - \frac{1}{3}x^3 - \frac{1}{4}x^4 - ..........$$
$$e^x = \sum_{n = 0}^\infty \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!}+ ....$$
$$\therefore y = \log(1+x)$$
$$x = -\frac{1}{1!}\cdot\frac{3}{4} + \frac{1}{2!}\cdot(\frac{3}{4})^2 -\frac{1}{3!}\cdot(\frac{3}{4})^3 +....$$
$$= e^{\frac{-3}{4}} - 1$$
$$\therefore y = \log(e^{\frac{-3}{4}})$$
$$= \frac{-3}{4}$$
Hence A is the correct answer.
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