The value of sumlimits_{n = 1}^infty {frac{{^ | vedclass

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(C) We know that the sum of binomial coefficients $\sum_{k=0}^{n} {^nC_k} = 2^n$ and the permutation $^nP_n = n!$. Substituting these into the given e

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$e^2 - 1$

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(C) We know that the sum of binomial coefficients $\sum_{k=0}^{n} {^nC_k} = 2^n$ and the permutation $^nP_n = n!$. Substituting these into the given expression: $\sum_{n=1}^{\infty} \frac{2^n}{n!} = \sum_{n=0}^{\infty} \frac{2^n}{n!} - \frac{2^0}{0!}$ Since the Taylor series expansion for $e^x$ is $\sum_{n=0}^{\infty} \frac{x^n}{n!}$,for $x=2$,we have $\sum_{n=0}^{\infty} \frac{2^n}{n!} = e^2$. Therefore,the sum is $e^2 - 1$.

The value of $\sum\limits_{n = 1}^\infty {\frac{{^n{C_0} + ... + ^n{C_n}}}{{^n{P_n}}}} $ is

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