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

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Question

(B) We know that $\log_a(1 + x) = \log_e(1 + x) \cdot \log_a e$. Using the standard logarithmic series expansion,$\log_e(1 + x) = \sum_{n=1}^{\infty}

Vedclass · Accepted Answer

$\frac{(-1)^{n-1}}{n} \log_a e$

Vedclass · Answer

(B) We know that $\log_a(1 + x) = \log_e(1 + x) \cdot \log_a e$. Using the standard logarithmic series expansion,$\log_e(1 + x) = \sum_{n=1}^{\infty} (-1)^{n-1} \frac{x^n}{n}$. Substituting this into the expression,we get: $\log_a(1 + x) = \log_a e \left( \sum_{n=1}^{\infty} (-1)^{n-1} \frac{x^n}{n} \right)$. Therefore,the coefficient of $x^n$ is $\frac{(-1)^{n-1}}{n} \log_a e$.

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

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