frac{1}{1!} + frac{4}{2!} + frac{7}{3!} + fra | vedclass

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(B) The general term of the series is $T_n = \frac{3n - 2}{n!}$.We can rewrite the general term as $T_n = \frac{3n}{n!} - \frac{2}{n!} = \frac{3}{(n -

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(B) The general term of the series is $T_n = \frac{3n - 2}{n!}$.We can rewrite the general term as $T_n = \frac{3n}{n!} - \frac{2}{n!} = \frac{3}{(n - 1)!} - \frac{2}{n!}$.The sum of the series is $S = \sum_{n=1}^{\infty} T_n = \sum_{n=1}^{\infty} \left( \frac{3}{(n - 1)!} - \frac{2}{n!} \right)$.$S = 3 \sum_{n=1}^{\infty} \frac{1}{(n - 1)!} - 2 \sum_{n=1}^{\infty} \frac{1}{n!}$.We know that $e = \sum_{n=0}^{\infty} \frac{1}{n!} = 1 + \frac{1}{1!} + \frac{1}{2!} + \dots$ and $\sum_{n=1}^{\infty} \frac{1}{(n - 1)!} = \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \dots = e$.Also,$\sum_{n=1}^{\infty} \frac{1}{n!} = e - 1$.Substituting these values,$S = 3(e) - 2(e - 1) = 3e - 2e + 2 = e + 2$.

$\frac{1}{1!} + \frac{4}{2!} + \frac{7}{3!} + \frac{10}{4!} + \dots \infty = $

Similar Questions

For every real number $x$, let $f(x) = \frac{x}{1!} + \frac{3}{2!} x^2 + \frac{7}{3!} x^3 + \frac{15}{4!} x^4 + \dots$. Then the equation $f(x) = 0$ has

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