[题目]证明:当 geqslant 0 时, (x)^n-1-(n-1)(x)^nleqslant 1,-|||-(正整数 gt 1 ).

【题目】-|||-1 0 1)-|||-设A= 0 2 0 而 geqslant 2 为正整数,则 ^n-2(A)^n-1=-|||-1 0 1

3.设实数 in (0,1) .数列(xn)满足 _(0)=1 ,且对任意正整数n,均有 _(n)=dfrac (1)({x)_(n-1)}+a 证明:对任-|

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

若方程a_0x^n+a_1x^n-1+…+a_(n-1)x=0有一个正根x=x_0, 证明方程a_0nx^n-1+a_1(n-1)x^n-2+…+a_(n-1)

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

14)设 (x)=arctan dfrac (1+x)(1-x), 整数 geqslant 0, 则 ^(2n+1)(0)= __

=(f)^(n-1)((x)_(0))=0.-|||-^(n)((x)_(0))neq 0, 证明:-|||-(1)当n为奇数时,f(x)在x0处不取得极值;-

+(a)_(n)=0, 求证:方程 (a)_(n)(x)^n-1+(n-1)(a)_(n-1)(x)^n-2+... +2(a)_(2)x+-|||-_(1)=