JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

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Application +4 | -1

{\log _e}\frac{{1 + 3x}}{{1 - 2x}}

is equal to

View answer and explanation

Correct option: C

5x - \frac{{5{x^2}}}{2} + \frac{{35{x^3}}}{3} - \; \ldots

Option D (

5x + \frac{{5{x^2}}}{2} + \frac{{35{x^3}}}{3} + \ldots

) is correct because expanding the logarithmic expression using the series expansion of log(1+x) and log(1-x) results in the given series.

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\frac{3}{4}