The coefficient of x^n in the expansion of lo | vedclass

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(B) We have $\log_e(1 + 3x + 2x^2) = \log_e((1 + x)(1 + 2x)) = \log_e(1 + x) + \log_e(1 + 2x)$.Using the expansion $\log_e(1 + y) = \sum_{n=1}^{\infty

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$\frac{(-1)^{n+1}}{n} [2^n + 1]$

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(B) We have $\log_e(1 + 3x + 2x^2) = \log_e((1 + x)(1 + 2x)) = \log_e(1 + x) + \log_e(1 + 2x)$.Using the expansion $\log_e(1 + y) = \sum_{n=1}^{\infty} (-1)^{n-1} \frac{y^n}{n}$,we get:$\log_e(1 + x) = \sum_{n=1}^{\infty} (-1)^{n-1} \frac{x^n}{n}$$\log_e(1 + 2x) = \sum_{n=1}^{\infty} (-1)^{n-1} \frac{(2x)^n}{n} = \sum_{n=1}^{\infty} (-1)^{n-1} \frac{2^n x^n}{n}$Adding these,the coefficient of $x^n$ is $(-1)^{n-1} \left( \frac{1}{n} + \frac{2^n}{n} \right) = (-1)^{n-1} \frac{2^n + 1}{n}$.Since $(-1)^{n-1} = (-1)^{n+1}$,the coefficient is $\frac{(-1)^{n+1}(2^n + 1)}{n}$.

The coefficient of $x^n$ in the expansion of $\log_e(1 + 3x + 2x^2)$ is

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