If {T_n} = frac{{{3^n}}}{{2(n!)}} - frac{1}{{ | vedclass

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(B) Given that ${T_n} = \frac{1}{2}\left[ {\frac{{{3^n}}}{{n!}} - \frac{1}{{n!}}} \right]$.Therefore,the sum of the series is ${S_\infty} = \sum_{n=1}

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$\frac{{{e^3} - e}}{2}$

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(B) Given that ${T_n} = \frac{1}{2}\left[ {\frac{{{3^n}}}{{n!}} - \frac{1}{{n!}}} \right]$.Therefore,the sum of the series is ${S_\infty} = \sum_{n=1}^\infty {T_n} = \frac{1}{2} \left[ \sum_{n=1}^\infty \frac{3^n}{n!} - \sum_{n=1}^\infty \frac{1}{n!} \right]$.We know that the exponential series is $e^x = \sum_{n=0}^\infty \frac{x^n}{n!} = 1 + \sum_{n=1}^\infty \frac{x^n}{n!}$,which implies $\sum_{n=1}^\infty \frac{x^n}{n!} = e^x - 1$.For the first part,$\sum_{n=1}^\infty \frac{3^n}{n!} = e^3 - 1$.For the second part,$\sum_{n=1}^\infty \frac{1^n}{n!} = e^1 - 1 = e - 1$.Substituting these values back into the expression for ${S_\infty}$:${S_\infty} = \frac{1}{2} \left[ (e^3 - 1) - (e - 1) \right] = \frac{1}{2} (e^3 - 1 - e + 1) = \frac{e^3 - e}{2}$.

If ${T_n} = \frac{{{3^n}}}{{2(n!)}} - \frac{1}{{2(n!)}}$,then ${S_\infty } = $

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