10.[单选题]-|||-10、高阶导数10-|||-((ln x))^(n)=-|||-A) ((-1))^n-1dfrac ((n-1)!)({x)^n-1}-|||-B ((-1))^n-1dfrac (n!)({x)^n}-|||-C dfrac ((n-1)!)({x)^n}-|||-D ((-1))^n-1dfrac ((n-1)!)({x)^n}

+(n-1)ln dfrac (n-1)(n)] = ____

,-|||-;-|||-(3) (x)_(1)+(n-1)(x)_(2)+... +2(x)_(n-1)+(x)_(n)=0

_(n)=dfrac (1)(2)((x)_(n-1)+dfrac ({a)^2}({x)_(n-1)}) n=1,2,···, 证明数列(xn)极限存在,并求

dfrac (1)(n-1)sum _(i=1)^n(({X)_(i)-overline (X))}^2 .-|||-n-|||-C. sqrt (dfrac

[单选题]y（n）+0.3y（n-1）=x（n）与y（n）=-0.2x（n）+x（n-1）是（）.

设x_(0)=0,x_(n)=(1+2x_(n-1))/(1+x_(n-1))(n=1,2,3,...),则lim_(ntoinfty)x_(n)=设$x_{0

+(a)_(n-1)x=0 有一个正根 =(x)_(0), 证明方程 _(0)n(x)^n-1+(a)_(1)(n-1)(x)^n-2+... +(a)_(n-

+(a)_(n-1)x=0 有一个正根 =(x)_(0) ,证明方程 _(0)n(x)^n-1+(a)_(1)(n-1)(x)^n-2+... +(a)_(n-

leqslant x(n) 为其次序统计量,令-|||-._(i)=dfrac ({x)_((i))}({x)_((i+1))} i=1 ,···, n-1 ,

4.样本X1,X2,···Xn来自总体 sim N(0,1) , overline (X)=dfrac (1)(n)sum _(i=1)^n(X)_(i) ,