∑n=1∞C0 n+C1 n+……+Cn nPn n=

∑n=1∞C0 n+C1 n+……+Cn nPn n =∑n=1∞ 2nn=e2-1

∑n=1∞C0 n+C1 n+……+Cn nPn n=

∑n=1∞C0 n+C1 n+……+Cn nPn n =∑n=1∞ 2nn=e2-1

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$\sum_{n = 1}^{\infty} \frac{{}^{n}C_{0} + {}^{n}C_{1} + \dots \dots + {}^{n}C_{n}}{{}^{n}P_{n}} =$

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$e^{2}$

$e^{2} - 1$

$e^{3}$

$e^{3} - 1$

answer is B.

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$\sum_{n = 1}^{\infty} \frac{{}^{n}C_{0} + {}^{n}C_{1} + \dots \dots + {}^{n}C_{n}}{{}^{n}P_{n}} = \sum_{n = 1}^{\infty} \frac{2^{n}}{n} = e^{2} - 1$

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∑n=1∞C0 n+C1 n+……+Cn nPn n=

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$\sum_{n = 1}^{\infty} \frac{{}^{n}C_{0} + {}^{n}C_{1} + \dots \dots + {}^{n}C_{n}}{{}^{n}P_{n}} =$