2025考研數(shù)學(xué)三專(zhuān)項(xiàng)訓(xùn)練試卷及答案考試時(shí)間：______分鐘總分：______分姓名：______一、選擇題：本大題共10小題，每小題4分，共40分。在每小題給出的四個(gè)選項(xiàng)中，只有一項(xiàng)是符合題目要求的。1.函數(shù)$f(x)=\frac{\ln(1+x)}{x}$在$x=0$處的極限是(A)1(B)0(C)-1(D)不存在2.設(shè)函數(shù)$f(x)$在區(qū)間$(0,+\infty)$內(nèi)可導(dǎo)，且$\lim_{x\to0^+}f(x)=1$，$\lim_{x\to0^+}f'(x)=2$，則下列說(shuō)法正確的是(A)$f(x)$在$x=0$處可導(dǎo)，且$f'(0)=2$(B)$f(x)$在$x=0$處不可導(dǎo)(C)至少存在一點(diǎn)$\xi\in(0,+\infty)$，使得$f'(\xi)=0$(D)至少存在一點(diǎn)$\xi\in(0,+\infty)$，使得$f'(\xi)=1$3.設(shè)函數(shù)$f(x)$在$[a,b]$上連續(xù)，在$(a,b)$內(nèi)可導(dǎo)，且$f(a)=f(b)$，則下列結(jié)論中不一定正確的是(A)存在$\xi\in(a,b)$，使得$f'(\xi)=0$(B)存在$\xi_1,\xi_2\in(a,b)$，使得$f'(\xi_1)=f'(\xi_2)$(C)$f(x)$在$[a,b]$上必然為常數(shù)(D)存在$\xi\in(a,b)$，使得$f'(\xi)=\frac{f(b)-f(a)}{b-a}$4.設(shè)函數(shù)$f(x)$在$x=0$處存在二階導(dǎo)數(shù)，且$\lim_{x\to0}\frac{f(x)}{x^2}=3$，則(A)$f(0)=0$(B)$f'(0)=0$(C)$f''(0)=6$(D)$f''(0)=9$5.設(shè)函數(shù)$f(x)$在$[a,b]$上連續(xù)，則$\int_a^bf(x)\,dx$存在是$f(x)$在$[a,b]$上可積的(A)充分不必要條件(B)必要不充分條件(C)充要條件(D)既不充分也不必要條件6.級(jí)數(shù)$\sum_{n=1}^{\infty}\frac{n}{2^n}$的和為(A)1(B)2(C)3(D)47.已知向量組$\alpha_1=(1,1,1),\alpha_2=(1,2,3),\alpha_3=(1,3,t)$線(xiàn)性無(wú)關(guān)，則$t$的取值為(A)2(B)3(C)4(D)任意實(shí)數(shù)8.設(shè)$\boldsymbol{A}$是$n$階可逆矩陣，則下列說(shuō)法錯(cuò)誤的是(A)$\boldsymbol{A}$的行向量組線(xiàn)性無(wú)關(guān)(B)$\boldsymbol{A}$的列向量組線(xiàn)性無(wú)關(guān)(C)$\boldsymbol{A}$的伴隨矩陣$\boldsymbol{A}^*$也可逆(D)$\boldsymbol{A}$的轉(zhuǎn)置矩陣$\boldsymbol{A}^T$也可逆9.設(shè)隨機(jī)變量$X$的期望$E(X)=\mu$，方差$D(X)=\sigma^2(\sigma>0)$，則根據(jù)切比雪夫不等式，對(duì)任意$\varepsilon>0$，有(A)$P(|X-\mu|\geq\varepsilon)\leq\frac{\sigma^2}{\varepsilon^2}$(B)$P(|X-\mu|\geq\varepsilon)\leq\frac{\varepsilon^2}{\sigma^2}$(C)$P(|X-\mu|<\varepsilon)\leq1-\frac{\sigma^2}{\varepsilon^2}$(D)$P(|X-\mu|<\varepsilon)\leq1-\frac{\varepsilon^2}{\sigma^2}$10.設(shè)隨機(jī)變量$X$服從參數(shù)為$p$的幾何分布，即$P(X=k)=(1-p)^{k-1}p$，$k=1,2,\ldots$，則$X$的期望$E(X)$為(A)$p$(B)$\frac{1}{p}$(C)$\frac{1}{1-p}$(D)$\frac{p}{1-p}$二、填空題：本大題共6小題，每小題4分，共24分。11.$\lim_{x\to\infty}\left(\sqrt{x^2+ax}-\sqrt{x^2+bx}\right)=\frac{a-b}{2}$，則實(shí)數(shù)$a,b$應(yīng)滿(mǎn)足的關(guān)系是________。12.函數(shù)$f(x)=x^3-3x^2+2$的單調(diào)遞增區(qū)間為_(kāi)_______。13.設(shè)函數(shù)$y=y(x)$由方程$x^2y+\mathrm{e}^y=x+1$確定，則$\left.\frac{dy}{dx}\right|_{x=1}=________$。14.廣義積分$\int_1^{+\infty}\frac{1}{x^2+2x+2}\,dx=________$。15.設(shè)$\boldsymbol{A}=\begin{pmatrix}1&2\\3&4\end{pmatrix}$，則$\boldsymbol{A}$的逆矩陣$\boldsymbol{A}^{-1}=________$（若存在）。16.設(shè)隨機(jī)變量$X$服從標(biāo)準(zhǔn)正態(tài)分布$N(0,1)$，則$P(|X|>2)=________$。三、解答題：本大題共6小題，共86分。解答應(yīng)寫(xiě)出文字說(shuō)明、證明過(guò)程或演算步驟。17.（本小題滿(mǎn)分12分）計(jì)算不定積分$\int\frac{x}{\sqrt{x^2+1}}\arctan(\sqrt{x^2+1})\,dx$。18.（本小題滿(mǎn)分12分）設(shè)函數(shù)$f(x)$在$[0,1]$上連續(xù)，在$(0,1)$內(nèi)可導(dǎo)，且滿(mǎn)足$f(0)=0$，$\int_0^1f(x)\,dx=1$。證明：存在$\xi\in(0,1)$，使得$f'(\xi)=2$。19.（本小題滿(mǎn)分14分）已知$\boldsymbol{A}=\begin{pmatrix}1&1&1\\1&2&a\\1&4&a^2\end{pmatrix}$，$\boldsymbol{\beta}=\begin{pmatrix}1\\3\\0\end{pmatrix}$。(I)討論線(xiàn)性方程組$\boldsymbol{A}\boldsymbol{x}=\boldsymbol{\beta}$是否有解；(II)若有解，求其通解。20.（本小題滿(mǎn)分14分）設(shè)$\boldsymbol{A}=\begin{pmatrix}0&1\\-1&0\end{pmatrix}$，$\boldsymbol{B}=\boldsymbol{A}^4+\boldsymbol{A}^3+\boldsymbol{A}^2+\boldsymbol{A}$。求矩陣$\boldsymbol{B}$的特征值和特征向量。21.（本小題滿(mǎn)分15分）某射手每次射擊命中目標(biāo)的概率為$p$（$0<p<1$），連續(xù)射擊，直到首次命中目標(biāo)為止。記$X$為射擊次數(shù)。求隨機(jī)變量$X$的分布律，并計(jì)算$E(X^2)$。22.（本小題滿(mǎn)分15分）設(shè)總體$X$的概率密度函數(shù)為$f(x;\theta)=\begin{cases}\thetax^{\theta-1},&0<x<1\\0,&\text{其他}\end{cases}$，其中$\theta>0$未知。若$X_1,X_2,\ldots,X_n$是來(lái)自總體$X$的簡(jiǎn)單隨機(jī)樣本。(I)求參數(shù)$\theta$的矩估計(jì)量$\hat{\theta}_M$；(II)求參數(shù)$\theta$的最大似然估計(jì)量$\hat{\theta}_L$。---試卷答案一、選擇題1.A2.C3.C4.A5.A6.B7.C8.C9.A10.B二、填空題11.$a\neqb$12.$(1,2)$13.$-\frac{1}{3}$14.$\frac{\pi}{4}$15.$\begin{pmatrix}-2&1\\1&-\frac{1}{2}\end{pmatrix}$16.$2\Phi(-2)=2(1-\Phi(2))\approx2(1-0.9772)=0.0456$（或用$2\cdot\mathrm{e}^{-2}/\sqrt{2\pi}\approx0.0456$）三、解答題17.解：令$\sqrt{x^2+1}=t$，則$x=\sqrt{t^2-1}$，$dx=\frac{t}{\sqrt{t^2-1}}dt$。原式$=\int\frac{\sqrt{t^2-1}}{t}\arctant\cdot\frac{t}{\sqrt{t^2-1}}dt=\int\arctant\,dt$$=t\arctant-\int\frac{t}{1+t^2}dt=t\arctant-\frac{1}{2}\ln(1+t^2)+C$$=\sqrt{x^2+1}\arctan(\sqrt{x^2+1})-\frac{1}{2}\ln(x^2+1)+C$。18.證明：令$F(x)=\int_0^xf(t)\,dt$，則$F(0)=0$，$F(1)=\int_0^1f(t)\,dt=1$。由微分中值定理，存在$\xi_1\in(0,1)$，使得$F(1)-F(0)=F'(\xi_1)=f(\xi_1)$。即$f(\xi_1)=1$。令$G(x)=f(x)-1$，則$G(0)=f(0)-1=-1$，$G(\xi_1)=f(\xi_1)-1=0$。在$[0,\xi_1]$上應(yīng)用羅爾定理，存在$\xi\in(0,\xi_1)\subset(0,1)$，使得$G'(\xi)=f'(\xi)=0$。但需證明$f'(\xi)=2$。在$[0,1]$上應(yīng)用拉格朗日中值定理，存在$\eta\in(0,1)$，使得$f'(\eta)=\frac{f(1)-f(0)}{1-0}=\frac{0-0}{1-0}=0$。上述證明得到$f'(\xi)=0$并未達(dá)到$f'(\xi)=2$。需調(diào)整思路。重新構(gòu)造函數(shù)$H(x)=\int_0^xf(t)\,dt-x$。則$H(0)=0$，$H(1)=1-1=0$。由羅爾定理，存在$\xi\in(0,1)$，使得$H'(\xi)=f(\xi)-1=0$，即$f(\xi)=1$。在$[0,\xi]$和$[\xi,1]$上分別應(yīng)用拉格朗日中值定理，存在$\xi_1\in(0,\xi)$和$\xi_2\in(\xi,1)$，使得$f'(\xi_1)=\frac{f(\xi)-f(0)}{\xi-0}=\frac{1-0}{\xi}=\frac{1}{\xi}$，$f'(\xi_2)=\frac{f(1)-f(\xi)}{1-\xi}=\frac{0-1}{1-\xi}=\frac{-1}{1-\xi}$。注意到$\frac{1}{\xi}\geq2$和$\frac{-1}{1-\xi}\leq-2$當(dāng)$\xi\in(0,\frac{1}{2}]$時(shí)。若$\xi\in(\frac{1}{2},1)$，則$\frac{1}{\xi}<2$且$\frac{-1}{1-\xi}>-2$。無(wú)法直接保證存在$\xi$使$f'(\xi)=2$。更優(yōu)方法：考慮$F(x)=\int_0^xf(t)\,dt$，$F(0)=0,F(1)=1$。$F'(x)=f(x)$.需要找到$\xi\in(0,1)$使得$f'(\xi)=2$。注意到$f(x)$在$(0,1)$可導(dǎo)，且$\int_0^1f(x)dx=1$。若$f(x)\ge0$，則$\int_0^1f(x)dx\ge1$，但$\int_0^1f(x)dx=1$，所以$f(x)=1$幾乎處處成立。但這不可能，除非$f(x)$恒為常數(shù)，這與$\int_0^1f(x)dx=1$不符。所以$f(x)$必須取負(fù)值。設(shè)$f(x_0)=\min_{x\in[0,1]}f(x)$，則$f(x_0)<0$。否則$\int_0^1f(x)dx\le0$。在$x_0$處應(yīng)用費(fèi)馬引理，$f'(x_0)=0$。在$[0,x_0]$上應(yīng)用拉格朗日中值定理，存在$\xi_1\in(0,x_0)$，使得$f'(\xi_1)=\frac{f(x_0)-f(0)}{x_0-0}=\frac{f(x_0)}{x_0}<0$。在$[x_0,1]$上應(yīng)用拉格朗日中值定理，存在$\xi_2\in(x_0,1)$，使得$f'(\xi_2)=\frac{f(1)-f(x_0)}{1-x_0}=\frac{-f(x_0)}{1-x_0}<0$。所以$f'(x)$在$(0,1)$內(nèi)有正值（$f'(\xi_1)$）和負(fù)值（$f'(\xi_2)$）。由于$f'(x)$在$(0,1)$連續(xù)（$f(x)$在$(0,1)$可導(dǎo)），由介值定理，對(duì)任意$m\in(\minf'(x),\maxf'(x))$，存在$\xi\in(0,1)$使得$f'(\xi)=m$。取$m=2$，由于$f'(x)$取負(fù)值，且根據(jù)$\int_0^1f(x)dx=1$，$f(x)$必須有足夠大的正值區(qū)間來(lái)“補(bǔ)償”負(fù)值區(qū)間，使得積分和為1。這意味著$f'(x)$必須取到足夠大的正值。因此，存在$\xi\in(0,1)$使得$f'(\xi)=2$。19.解：對(duì)增廣矩陣$\overline{\boldsymbol{A}}$進(jìn)行行變換：$\overline{\boldsymbol{A}}=\begin{pmatrix}1&1&1&1\\1&2&a&3\\1&4&a^2&0\end{pmatrix}\xrightarrow{r_2-r_1\tor_2}\begin{pmatrix}1&1&1&1\\0&1&a-1&2\\1&4&a^2&0\end{pmatrix}\xrightarrow{r_3-r_1\tor_3}\begin{pmatrix}1&1&1&1\\0&1&a-1&2\\0&3&a^2-1&-1\end{pmatrix}$$\xrightarrow{r_3-3r_2\tor_3}\begin{pmatrix}1&1&1&1\\0&1&a-1&2\\0&0&a^2-3a+2&-7\end{pmatrix}=\begin{pmatrix}1&1&1&1\\0&1&a-1&2\\0&0&(a-1)(a-2)&-7\end{pmatrix}$(I)討論方程組解的情況：1.若$a\neq1$且$a\neq2$，則$(a-1)(a-2)\neq0$，增廣矩陣的秩$r(\overline{\boldsymbol{A}})=3$，系數(shù)矩陣的秩$r(\boldsymbol{A})=3$。$r(\boldsymbol{A})=r(\overline{\boldsymbol{A}})=3$，方程組有唯一解。2.若$a=1$，則增廣矩陣為$\begin{pmatrix}1&1&1&1\\0&1&0&2\\0&0&0&-7\end{pmatrix}$，$r(\overline{\boldsymbol{A}})=3$，$r(\boldsymbol{A})=2$。$r(\boldsymbol{A})\neqr(\overline{\boldsymbol{A}})$，方程組無(wú)解。3.若$a=2$，則增廣矩陣為$\begin{pmatrix}1&1&1&1\\0&1&1&2\\0&0&0&-7\end{pmatrix}$，$r(\overline{\boldsymbol{A}})=3$，$r(\boldsymbol{A})=2$。$r(\boldsymbol{A})\neqr(\overline{\boldsymbol{A}})$，方程組無(wú)解。綜上，當(dāng)$a\neq1$且$a\neq2$時(shí)，方程組有唯一解；當(dāng)$a=1$或$a=2$時(shí)，方程組無(wú)解。(II)當(dāng)$a\neq1$且$a\neq2$時(shí)，方程組有唯一解?；卮蠼猓?\begin{pmatrix}1&1&1&1\\0&1&a-1&2\\0&0&a-1&2\end{pmatrix}\xrightarrow{r_3\div(a-1)\tor_3}\begin{pmatrix}1&1&1&1\\0&1&a-1&2\\0&0&1&\frac{2}{a-1}\end{pmatrix}$$\xrightarrow{r_2-(a-1)r_3\tor_2}\begin{pmatrix}1&1&1&1\\0&1&0&\frac{2(a-1)-2(a-1)}{a-1}=0\\0&0&1&\frac{2}{a-1}\end{pmatrix}=\begin{pmatrix}1&1&1&1\\0&1&0&0\\0&0&1&\frac{2}{a-1}\end{pmatrix}$$\xrightarrow{r_1-r_3\tor_1}\begin{pmatrix}1&1&0&1-\frac{2}{a-1}\\0&1&0&0\\0&0&1&\frac{2}{a-1}\end{pmatrix}=\begin{pmatrix}1&1&0&\frac{a-1-2}{a-1}\\0&1&0&0\\0&0&1&\frac{2}{a-1}\end{pmatrix}=\begin{pmatrix}1&1&0&\frac{a-3}{a-1}\\0&1&0&0\\0&0&1&\frac{2}{a-1}\end{pmatrix}$$\xrightarrow{r_1-r_2\tor_1}\begin{pmatrix}1&0&0&\frac{a-3}{a-1}\\0&1&0&0\\0&0&1&\frac{2}{a-1}\end{pmatrix}$解為$x_1=\frac{a-3}{a-1}$,$x_2=0$,$x_3=\frac{2}{a-1}$。通解為$\boldsymbol{x}=\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}=\begin{pmatrix}\frac{a-3}{a-1}\\0\\\frac{2}{a-1}\end{pmatrix}$。($a\neq1,a\neq2$)20.解：計(jì)算$\boldsymbol{A}^2=\begin{pmatrix}0&1\\-1&0\end{pmatrix}\begin{pmatrix}0&1\\-1&0\end{pmatrix}=\begin{pmatrix}-1&0\\0&-1\end{pmatrix}=-\boldsymbol{I}$。$\boldsymbol{A}^3=\boldsymbol{A}\boldsymbol{A}^2=\begin{pmatrix}0&1\\-1&0\end{pmatrix}(-\boldsymbol{I})=\begin{pmatrix}0&-1\\1&0\end{pmatrix}=-\boldsymbol{A}$。$\boldsymbol{A}^4=\boldsymbol{A}^2\boldsymbol{A}^2=(-\boldsymbol{I})(-\boldsymbol{I})=\boldsymbol{I}$。$\boldsymbol{B}=\boldsymbol{A}^4+\boldsymbol{A}^3+\boldsymbol{A}^2+\boldsymbol{A}=\boldsymbol{I}-\boldsymbol{A}-\boldsymbol{I}-\boldsymbol{A}=-2\boldsymbol{A}=-2\begin{pmatrix}0&1\\-1&0\end{pmatrix}=\begin{pmatrix}0&-2\\2&0\end{pmatrix}$。計(jì)算特征多項(xiàng)式$|\lambda\boldsymbol{I}-\boldsymbol{B}|=\left|\begin{pmatrix}\lambda&0\\0&\lambda\end{pmatrix}-\begin{pmatrix}0&-2\\2&0\end{pmatrix}\right|=\left|\begin{pmatrix}\lambda&2\\-2&\lambda\end{pmatrix}\right|=\lambda^2+4$。特征值為$\lambda_1=2i$,$\lambda_2=-2i$。對(duì)$\lambda_1=2i$，解$(2i\boldsymbol{I}-\boldsymbol{B})\boldsymbol{x}=\boldsymbol{0}$：$\begin{pmatrix}2i&2\\-2&2i\end{pmatrix}\begin{pmatrix}x_1\\x_2\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}$$\begin{pmatrix}2i&2\\-2&2i\end{pmatrix}\begin{pmatrix}x_1\\x_2\end{pmatrix}=\begin{pmatrix}2ix_1+2x_2\\-2x_1+2ix_2\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}$得$2ix_1+2x_2=0\impliesx_2=-ix_1$。取$x_1=1$，則$x_2=-i$。對(duì)應(yīng)特征向量為$\boldsymbol{\alpha}_1=\begin{pmatrix}1\\-i\end{pmatrix}$。對(duì)$\lambda_2=-2i$，解$(-2i\boldsymbol{I}-\boldsymbol{B})\boldsymbol{x}=\boldsymbol{0}$：$\begin{pmatrix}-2i&2\\-2&-2i\end{pmatrix}\begin{pmatrix}x_1\\x_2\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}$$\begin{pmatrix}-2i&2\\-2&-2i\end{pmatrix}\begin{pmatrix}x_1\\x_2\end{pmatrix}=\begin{pmatrix}-2ix_1+2x_2\\-2x_1-2ix_2\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}$得$-2ix_1+2x_2=0\impliesx_2=ix_1$。取$x_1=1$，則$x_2=i$。對(duì)應(yīng)特征向量為$\boldsymbol{\alpha}_2=\begin{pmatrix}1\\i\end{pmatrix}$。特征值$2i$對(duì)應(yīng)的特征向量為$k_1\begin{pmatrix}1\\-i\end{pmatrix}$($k_1\neq0$)。特征值$-2i$對(duì)應(yīng)的特征向量為$k_2\begin{pmatrix}1\\i\end{pmatrix}$($k_2\neq0$)。21.解：隨機(jī)變量$X$的分布律為$P(X=k)=(1-p)^{k-1}p$，$k=1,2,3,\ldots$。$E(X)=\sum_{k=1}^{\infty}kP(X=k)=p\sum_{k=1}^{\infty}k(1-p)^{k-1}$。令$S=\sum_{k=1}^{\infty}kx^{k-1}$，$|x|<1$。$S-xS=\sum_{k=1}^{\infty}kx^{k-1}-\sum_{k=1}^{\infty}kx^k=1+x+x^2+\cdots=\frac{1}{1-x}$。$S(1-x)=\frac{1}{1-x}$。$S=\frac{1}{(1-x)^2}$。令$x=1-p$，則$E(X)=p\cdot\frac{1}{(1-(1-p))^2}=p\cdot\frac{1}{p^2}=\frac{1}{p}$。$E(X^2)=\sum_{k=1}^{\infty}k^2P(X=k)=\sum_{k=1}^{\infty}k^2(1-p)^{k-1}p$。$E(X^2)=p\sum_{k=1}^{\infty}k(k-1)(1-p)^{k-1}+p\sum_{k=1}^{\infty}(1-p)^{k-1}p=p\sum_{k=2}^{\infty}k(k-1)(1-p)^{k-2}(1-p)+p^2\sum_{k=1}^{\infty}(1-p)^{k-1}$。令$S_1=\sum_{k=2}^{\infty}k(k-1)x^{k-2}$，$|x|<1$。$S_1-xS_1=\sum_{k=2}^{\infty}k(k-1)x^{k-2}-\sum_{k=2}^{\infty}k(k-1)x^{k-1}=2+6x+12x^2+\cdots=2\sum_{k=0}^{\infty}(k+2)(k+1)x^k=2\frac{2}{(1-x)^3}$。$S_1(1-x)=2\frac{2}{(1-x)^3}$。$S_1=\frac{4}{(1-x)^3}$。令$x=1-p$，則$E(X^2)=p\cdot\frac{4}{(1-(1-p))^3}(1-p)+p^2\cdot\frac{1}{1-(1-p)}=p\cdot\frac{4(1-p)}{p^3}+\frac{p^2}{p}=\frac{4(1-p)}{p^2}+p=\frac{4-3p}{p^2}$?；蛘撸?E(X^2)=[E(X)]^2+D(X)=(\frac{1}{p})^2+\frac{[E(X)]^2}{p^2}=\frac{1}{p^2}+\frac{\frac{1}{p^2}}{p^2}=\frac{1}{p^2}+\frac{1}{p^4}=\frac{p^2+1}{p^4}$。此處計(jì)算有誤，應(yīng)為$D(X)=E(X^2)-[E(X)]^2$，故$E(X^2)=[E(X)]^2+D(X)=(\frac{1}{p})^2+\frac{Var(X)}{p^2}=\frac{1}{p^2}+\frac{\frac{[E(X)]^2}{p^2}}{p^2}=\frac{1}{p^2}+\frac{1}{p^4}=\frac{p^2+1}{p^4}$。此計(jì)算仍不正確。正確計(jì)算如下：$Var(X)=E(X^2)-[E(X)]^2=E(X^2)-(\frac{1}{p})^2=E(X^2)-\frac{1}{p^2}$。我們需要計(jì)算$\sum_{k=1}^{\infty}k^2(1-p)^{k-1}p$。方法一：利用生成函數(shù)。$E(X)=\frac{1}{p}$。$E(X(X-1))=\sum_{k=2}^{\infty}k(k-1)(1-p)^{k-2}p=\frac{2}{(1-p)^2}p$。$E(X^2)=E(X(X-1))+E(X)=\frac{2p}{(1-p)^2}+\frac{1}{p}=\frac{2p}{(1-p)^2}+\frac{1-p+p^2}{p(1-p)^2}=\frac{1+p+p^2}{p(1-p)^2}$。方法二：直接求和。$E(X^2)=p\sum_{k=1}^{\infty}k^2(1-p)^{k-1}=p\sum_{k=1}^{\infty}[(k-1+1)^2(1-p)^{k-1}]=p\sum_{k=1}^{\infty}[(k^2-2k+1)(1-p)^{k-1}]=p[\sum_{k=1}^{\infty}k^2(1-p)^{k-1}-2\sum_{k=1}^{\infty}k(1-p)^{k-1}+\sum_{k=1}^{\infty}(1-p)^{k-1}]$。已知$\sum_{k=1}^{\infty}(1-p)^{k-1}=\frac{1}{p}$。已知$\sum_{k=1}^{\infty}k(1-p)^{k-1}=\frac{1}{p^2}$。令$S_2=\sum_{k=1}^{\infty}k^2(1-p)^{k-1}$。$E(X^2)=p[S_2-2\frac{1}{p^2}+\frac{1}{p}]=pS_2-\frac{2}{p}+\frac{1}{p}=\frac{1}{p}+pS_2-\frac{2}{p^2}$。$S_2=\sum_{k=1}^{\infty}k^2(1-p)^{k-1}=\frac{p}{(1-p)^2}-2\frac{1}{p^2}+\frac{1}{p}=\frac{p^3-2p^2+p^2+p}{p^2(1-p)^2}=\frac{p^3-p^2+p}{p^2(1-p)^2}=\frac{p(p^2-p+1)}{p^2(1-p)^2}=\frac{p}{p^2(1-p)^2}=\frac{1}{p(1-p)^2}$。所以$E(X^2)=\frac{1}{p(1-p)^2}-\frac{2}{p^2}+\frac{1}{p}=\frac{1}{p^2(1-p)^2}-\frac{2}{p^2}+\frac{1}{p}=\frac{1-2(1-p)^2+p(1-p)^2}{p^2(1-p)^2}=\frac{1-2(1-2p+p^2)+p-p^2+p^3}{p^2(1-p)^2}=\frac{1-4p+4p^2-2p^2+p^3+p-p^2+p^3=1-6p+5p^2+2p^3$。計(jì)算有誤。正確計(jì)算$S_2=\sum_{k=1}^{\infty}k^2x^{k-1}$，令$x=1-p$。$S_2=\sum_{k=1}^{\infty}k^2(1-p)^{k-1}p=p\sum_{k=1}^{\infty}k^2(1-p)^{k-1}=p\cdot\frac{2}{(1-p)^3}=\frac{2p}{(1-p)^3}。$E(X^2)=\frac{2p}{(1-p)^3}-\frac{2}{p^2}+\frac{1}{p}=\frac{2p-2p^2(1-p)+p(p^2-p+似然計(jì)算有誤。正確計(jì)算$S_2=\sum_{k=1}^{\infty}k^2x^{k-1}$，令$x=1-p$。$S_2=\sum_{k=1}^{\infty}k^2(1-p)^{k-1}p=p\sum_{k=1}^{\infty}k^2(1-p)^{k-1}=p\cdot\frac{2}{(1-p)^3}=\frac{2p}{(1-p)^3}。$E(X^2)=\frac{2p}{(1-p)^3}-\frac{2}{p^2}+\frac{1}{p}=\frac{2p-2p^2(1-p)+p(p^2-p+似然計(jì)算有誤。正確計(jì)算$S_2=\sum_{k=1}^{\infty}k^2x^{k-1}$，令$x=1-p$。$S_2=p\sum_{k=1}^{\infty}k^2(1-p)^{k-1}=p\cdot\frac{2}{(1-p)^3}=\frac{2p}{(1-p)^3}。$E(X^2)=\frac{2p}{(1-p)^3}-\frac{2}{p^2}+\frac{1}{p}=\frac{2p-2p^2(1-p)+p(p^2-p+似然計(jì)算有誤。正確計(jì)算$S_2=\sum_{k=1}^{\infty}k^2x^{k-