(1 + 3)log_e 3 + frac{1 + 3^2}{2!} (log_e 3)^ | vedclass

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Question

(A) The given series is $S = \sum_{n=1}^{\infty} \frac{1 + 3^n}{n!} (\log_e 3)^n$. We can split this into two separate series: $S = \sum_{n=1}^{\infty

Vedclass · Accepted Answer

$28$

Vedclass · Answer

(A) The given series is $S = \sum_{n=1}^{\infty} \frac{1 + 3^n}{n!} (\log_e 3)^n$. We can split this into two separate series: $S = \sum_{n=1}^{\infty} \frac{(\log_e 3)^n}{n!} + \sum_{n=1}^{\infty} \frac{(3 \log_e 3)^n}{n!}$. Using the exponential series expansion $e^x = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \dots$,we know that $\sum_{n=1}^{\infty} \frac{x^n}{n!} = e^x - 1$. For the first part,$x = \log_e 3$,so $\sum_{n=1}^{\infty} \frac{(\log_e 3)^n}{n!} = e^{\log_e 3} - 1 = 3 - 1 = 2$. For the second part,$x = 3 \log_e 3 = \log_e 3^3 = \log_e 27$,so $\sum_{n=1}^{\infty} \frac{(\log_e 27)^n}{n!} = e^{\log_e 27} - 1 = 27 - 1 = 26$. Therefore,$S = 2 + 26 = 28$.

$(1 + 3)\log_e 3 + \frac{1 + 3^2}{2!} (\log_e 3)^2 + \frac{1 + 3^3}{3!} (\log_e 3)^3 + \dots \infty = $

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