Q. The expansion of $\log{(1 + 3x + 2x^{2})}$ is

A) $3x - \frac{5}{4}\ x^{2} + \frac{9}{3}x^{3} - \frac{17}{4}x^{4} + ...\infty$
B) $4x - \frac{5}{4}x^{2} + \frac{9}{3}x^{3} - \frac{17}{4}x^{4} + ...\infty$
C) $3x - \frac{5}{2}x^{2} + \frac{9}{3}x^{3} - \frac{17}{4}x^{4} + ...\infty$
D) $- 3x - \frac{5}{4}x^{2} - \frac{9}{3}x^{3} - \frac{17}{4}x^{4} - ...\infty$
Explanation

(c)
Given, $\log{(1 + 3x + 2x^{2})}$
$= \log(1 + x) - \log(1 + 2x)$
$= \left( x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \frac{x^{4}}{4} + ... \right) + \left( 2x - \frac{(2x)^{2}}{2} + \frac{(2x)^{3}}{3} - ... \right)$
$= 3x - \frac{5}{2}x^{2} + \frac{9}{3}x^{3} - \frac{17}{4}x^{4} + ...$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.