(C) We know that $1+x+x^2 = \frac{1-x^3}{1-x}$.Thus,$\log(1+x+x^2) = \log(1-x^3) - \log(1-x)$.Using the expansion $\log(1-t) = -(t + \frac{t^2}{2} + \

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(C) We know that $1+x+x^2 = \frac{1-x^3}{1-x}$.Thus,$\log(1+x+x^2) = \log(1-x^3) - \log(1-x)$.Using the expansion $\log(1-t) = -(t + \frac{t^2}{2} + \frac{t^3}{3} + \dots)$,we get:$\log(1-x^3) = -(x^3 + \frac{x^6}{2} + \dots)$$-\log(1-x) = x + \frac{x^2}{2} + \frac{x^3}{3} + \dots$Adding these,the expansion is $x + \frac{x^2}{2} + (\frac{1}{3} - 1)x^3 + \dots$The coefficient of $x^3$ is $\frac{1}{3} - 1 = -\frac{2}{3}$.

Class 11 Mathematics Exponential and Logarithmic Series Logarithmic series

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For $|x| < 1$,the coefficient of $x^3$ in the expansion of $\log(1+x+x^2)$ in ascending powers of $x$ is:

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A
$\frac{2}{3}$
B
$\frac{4}{3}$
C
$-\frac{2}{3}$
D
$-\frac{4}{3}$

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Exponential and Logarithmic Series

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