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

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(B) The given series is of the form $\sum_{n=1}^{\infty} \frac{x^n}{n!}$,where $x = 2 \log_e 2$.We know that the expansion of $e^x$ is $1 + \frac{x}{1

Vedclass · Accepted Answer

$3$

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(B) The given series is of the form $\sum_{n=1}^{\infty} \frac{x^n}{n!}$,where $x = 2 \log_e 2$.We know that the expansion of $e^x$ is $1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots \infty$.Therefore,$\sum_{n=1}^{\infty} \frac{x^n}{n!} = e^x - 1$.Substituting $x = 2 \log_e 2 = \log_e(2^2) = \log_e 4$,we get:Sum $= e^{\log_e 4} - 1$.Since $e^{\log_e a} = a$,the sum is $4 - 1 = 3$.

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

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