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$\frac{1}{k!} \sum_{i = 2}^{k} (A_{k}^{i} - C_{k}^{i}) (\frac{3}{4})^{i} = \frac{1}{k!} \sum_{i = 2}^{k} (C_{k}^{i} i! - C_{k}^{i}) (\frac{3}{4})^{i} = \frac{1}{k!} \sum_{i = 2}^{k} C_{k}^{i} i! (\frac{3}{4})^{i} - \frac{1}{k!} \sum_{i = 2}^{k} C_{k}^{i} (\frac{3}{4})^{i} \frac{1}{k!}\sum_{i=2}^k(A_k^i-C_k^i)(\frac{3}{4})^i=\frac{1}{k!}\sum_{i=2}^k(C_k^ii!-C_k^i)(\frac{3}{4})^i=\frac{1}{k!}\sum_{i=2}^kC_k^ii!(\frac{3}{4})^i-\frac{1}{k!}\sum_{i=2}^kC_k^i(\frac{3}{4})^i$ k!1∑i=2k(Aki−Cki)(43)i=k!1∑i=2k(Ckii!−Cki)(43)i=k!1∑i=2kCkii!(43)i−k!1∑i=2kCki(43)i，对上两项分开求和

1.求 $\frac{1}{k!} \sum_{i = 2}^{k} C_{k}^{i} i! (\frac{3}{4})^{i} \frac{1}{k!}\sum_{i=2}^kC_k^ii!(\frac{3}{4})^i$ k!1∑i=2kCkii!(43)i，因为 $C_{k}^{i} = \frac{k!}{i! (k - i)!} C_k^i=\frac{k!}{i!(k-i)!}$ Cki=i!(k−i)!k!，代入得 $\frac{1}{k!} \sum_{i = 2}^{k} C_{k}^{i} i! (\frac{3}{4})^{i} = (\frac{3}{4})^{k} + \sum_{i = 2}^{k - 1} \frac{1}{(k - i)!} (\frac{3}{4})^{i} = (\frac{3}{4})^{k} + (\frac{3}{4})^{k} \sum_{i = 2}^{k - 1} \frac{1}{(k - i)!} (\frac{4}{3})^{(k - i)} \frac{1}{k!}\sum_{i=2}^kC_k^ii!(\frac{3}{4})^i=(\frac{3}{4})^k+\sum_{i=2}^{k-1}\frac{1}{(k-i)!}(\frac{3}{4})^i=(\frac{3}{4})^k+(\frac{3}{4})^k\sum_{i=2}^{k-1}\frac{1}{(k-i)!}(\frac{4}{3})^{(k-i)}$ k!1∑i=2kCkii!(43)i=(43)k+∑i=2k−1(k−i)!1(43)i=(43)k+(43)k∑i=2k−1(k−i)!1(34)(k−i)，又因为 $\lim_{k \to + \infty} \sum_{i = 2}^{k - 1} \frac{1}{(k - i)!} (\frac{4}{3})^{(k - i)} = e^{4 / 3} \lim_{k \to +\infty}\sum_{i=2}^{k-1}\frac{1}{(k-i)!}(\frac{4}{3})^{(k-i)}=e^{4/3}$ limk→+∞∑i=2k−1(k−i)!1(34)(k−i)=e4/3，于是原式 $= \lim_{k \to + \infty} (\frac{3}{4})^{k} + (\frac{3}{4})^{k} e^{4 / 3} = 0 =\lim_{k \to +\infty}(\frac{3}{4})^k+(\frac{3}{4})^ke^{4/3}=0$ =limk→+∞(43)k+(43)ke4/3=0

2.求 $\frac{1}{k!} \sum_{i = 2}^{k} C_{k}^{i} (\frac{3}{4})^{i} \frac{1}{k!}\sum_{i=2}^kC_k^i(\frac{3}{4})^i$ k!1∑i=2kCki(43)i，因为 $\sum_{i = 0}^{k} C_{k}^{i} (\frac{3}{4})^{i} = (1 + \frac{3}{4})^{k} \sum_{i=0}^kC_k^i(\frac{3}{4})^i=(1+\frac{3}{4})^k$ ∑i=0kCki(43)i=(1+43)k，所以原式 $= \lim_{k \to + \infty} \frac{(1 + \frac{3}{4})^{k} - 1 - \frac{3}{4} k}{k!} = 0 =\lim_{k \to +\infty}\frac{(1+\frac{3}{4})^k-1-\frac{3}{4}k}{k!}=0$ =limk→+∞k!(1+43)k−1−43k=0

综合式1，2，原式等于0

文章最后发布于: 2018-11-22 19:22:55

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