优化版

$\mathrm{赢}\mathrm{下}k\mathrm{次}\mathrm{游}\mathrm{戏}\mathrm{的}\mathrm{代}\mathrm{价}\mathrm{期}\mathrm{望}\mathrm{：}赢下\backslash \; k\backslash \; 次游戏的代价期望：$

$\mathrm{我}\mathrm{们}\mathrm{将}\mathrm{游}\mathrm{戏}\mathrm{在}\mathrm{第}i\mathrm{轮}\mathbf{\text{恰好结束}}\mathrm{的}\mathrm{概}\mathrm{率}\mathrm{记}\mathrm{为}D(i),\mathrm{且}i=k,k+1,k+2,\mathrm{.}\mathrm{.}\mathrm{.},\mathrm{\infty }我们将游戏在第\backslash \; i\backslash \; 轮\; \backslash textbf\{恰好结束\}\; 的概率记为\backslash \; D(i)\backslash \; ,且\backslash \; i=k,k+1,k+2,...,\backslash infty\backslash \backslash$

$\mathrm{因}\mathrm{为}\mathrm{要}\mathrm{恰}\mathrm{好}\mathrm{在}\mathrm{第}i\mathrm{轮}\mathrm{结}\mathrm{束}\mathrm{，}\mathrm{在}\mathrm{前}i-1\mathrm{轮}\mathrm{中}\mathrm{一}\mathrm{定}\mathrm{赢}\mathrm{过}k-1\mathrm{局}\mathrm{，}\mathrm{然}\mathrm{后}\mathrm{第}i\mathrm{轮}\mathrm{要}\mathrm{赢}因为要恰好在第i轮结束，在前\backslash \; i-1\backslash \; 轮中一定赢过\backslash \; k-1\backslash \; 局，然后第\backslash \; i\backslash \; 轮要赢\backslash \backslash$

$\mathrm{那}\mathrm{么}那么$

$D(i)=(\left(\genfrac{}{}{0px}{}{i-1}{k-1}\right)p^{k-1}(1-p{)}^{i-k})\times (p),(\mathrm{前}\mathrm{半}\mathrm{部}\mathrm{分}\mathrm{是}\mathrm{前}i-1\mathrm{轮}\mathrm{赢}k-1\mathrm{轮}\mathrm{的}\mathrm{概}\mathrm{率})D(i)=(\backslash binom\{i-1\}\{k-1\}p^\{k-1\}(1-p)^\{i-k\}\backslash \; )\backslash times(p),\backslash \; \backslash \; (前半部分是前\backslash \; i-1\backslash \; 轮赢k-1轮的概率)$

$\mathrm{在}\mathrm{第}i\mathrm{轮}\mathrm{游}\mathrm{戏}\mathrm{恰}\mathrm{好}\mathrm{结}\mathrm{束}\mathrm{的}\mathrm{代}\mathrm{价}\mathrm{是}\mathrm{显}\mathrm{然}\mathrm{的}\mathrm{，}\mathrm{为}({\sum }_{j=1}^{i}j)\mathrm{，}\mathrm{即}\frac{i(i+1)}{2}在第\backslash \; i\backslash \; 轮游戏恰好结束的代价是显然的，为\backslash \; (\backslash sum\_\{j=1\}^i\; j\backslash \; )\backslash \; ，即\backslash \; \backslash frac\{i(i+1)\}\{2\}\backslash \; \backslash \backslash$

$\mathrm{那}\mathrm{么}\mathrm{期}\mathrm{望}\mathrm{代}\mathrm{价}E\mathrm{也}\mathrm{就}\mathrm{出}\mathrm{来}\mathrm{了}\mathrm{，}E={\sum }_{i=k}^{\mathrm{\infty }}(\frac{i(i+1)}{2}\times D(i))那么期望代价\backslash \; E\backslash \; 也就出来了，E=\backslash sum\_\{i=k\}^\backslash infty\; (\backslash frac\{i(i+1)\}\{2\}\backslash times\; D(i))\backslash \backslash$

$\mathrm{即}E={\sum }_{i=k}^{\mathrm{\infty }}(\frac{i(i+1)}{2}\times \left(\genfrac{}{}{0px}{}{i-1}{k-1}\right)(1-p{)}^{i-k}{p}^k)即\backslash \; E=\backslash sum\_\{i=k\}^\backslash infty\; (\backslash \; \backslash frac\{i(i+1)\}\{2\}\backslash times\; \backslash binom\{i-1\}\{k-1\}(1-p)^\{i-k\}\backslash \; p^\{k\}\backslash \; )$

$\mathrm{注}\mathrm{意}\left(\genfrac{}{}{0px}{}{i-1}{k-1}\right)=\frac{(i-1)!}{(k-1)!(i-k)!}注意\backslash \; \backslash \; \backslash binom\{i-1\}\{k-1\}=\backslash frac\{(i-1)!\}\{(k-1)!\backslash \; (i-k)!\}$

$\mathrm{提}\mathrm{出}i\mathrm{的}\mathrm{无}\mathrm{关}\mathrm{项}\frac{p^k}{2(k-1)!},\mathrm{令}E=\frac{p^k}{2(k-1)!}f(1-p),\mathrm{此}\mathrm{处}提出i的无关项\backslash \; \backslash frac\{\backslash \; p^\{k\}\}\{2(k-1)!\}\backslash \; ,令\backslash \; E=\backslash frac\{\backslash \; p^\{k\}\}\{2(k-1)!\}f(1-p)\backslash \; ,\backslash \; 此处\backslash \backslash$

$f(x)={\sum }_{i=k}^{\mathrm{\infty }}({\prod }_{j=1}^{k+1}(i-k+j))x^{i-k}={\sum }_{i=0}^{\mathrm{\infty }}({\prod }_{j=1}^{k+1}(i-k+j))x^{i-k}f(x)=\backslash sum\_\{i=k\}^\backslash infty(\backslash \; \backslash prod\_\{j=1\}^\{k+1\}(i-k+j)\backslash \; )x^\{i-k\}\; =\backslash \; \backslash sum\_\{i=0\}^\backslash infty(\backslash \; \backslash prod\_\{j=1\}^\{k+1\}(i-k+j)\backslash \; )x^\{i-k\}$

$\mathrm{不}\mathrm{难}\mathrm{察}\mathrm{觉}f(x)\mathrm{与}\mathrm{幂}\mathrm{函}\mathrm{数}\mathrm{求}\mathrm{导}\mathrm{极}\mathrm{其}\mathrm{相}\mathrm{似}\mathrm{，}\mathrm{我}\mathrm{们}\mathrm{直}\mathrm{接}\mathrm{对}\mathrm{其}\mathrm{积}\mathrm{分}k+1\mathrm{次}不难察觉\backslash \; f(x)\backslash \; 与幂函数求导极其相似，我们直接对其积分k+1次$

$F(x)={\sum }_{i=0}^{\mathrm{\infty }}{x}^{i+1}={\sum }_{i=0}^{\mathrm{\infty }}{x}^i-1=\frac{1}{1-x}-1F(x)=\backslash sum\_\{i=0\}^\backslash infty\; x^\{i+1\}=\backslash sum\_\{i=0\}^\backslash infty\; x^\{i\}\backslash \; -1=\backslash frac\{1\}\{1-x\}\backslash \; -1$

$\mathrm{故}f(x)=(F(x){)}^{(k+1)}=(\frac{1}{1-x}-1{)}^{(k+1)}=\frac{(k+1)!}{(1-x{)}^{k+2}}故\backslash \; f(x)=(F(x))^\{(k+1)\}=(\backslash frac\{1\}\{1-x\}\backslash \; -1)^\{(k+1)\}=\backslash frac\{(k+1)!\}\{(1-x)^\{k+2\}\}$

$\mathrm{代}\mathrm{入}\mathrm{原}\mathrm{式}E=\frac{p^k}{2(k-1)!}f(1-p)=\frac{p^k}{2(k-1)!}\times \frac{(k+1)!}{(p{)}^{k+2}}=\frac{k(k+1)}{2p^2}代入原式\backslash \; E=\backslash frac\{\backslash \; p^\{k\}\}\{2(k-1)!\}f(1-p)=\backslash frac\{\backslash \; p^\{k\}\}\{2(k-1)!\}\backslash times\; \backslash frac\{(k+1)!\}\{(p)^\{k+2\}\}=\backslash frac\{k(k+1)\}\{2p^\{2\}\}$

$\mathrm{即}\mathrm{期}\mathrm{望}E=\frac{k(k+1)}{2p^2}即期望\backslash \; E=\backslash frac\{k(k+1)\}\{2p^\{2\}\}$