设λ1,λ2,···,λn-|||-是n阶方阵的特征值,则有-|||-sum _(i=1)^n(lambda )_(i)=sum _(i=1)^n(a)_(in)=(t)_(r)(A),-|||-这个正确还是错误。-|||-A.对-|||-B.错

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

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

13.设 sum _(i=1)^infty (a)_(n)=1, 则 sum _(n=1)^infty ((a)_(n)-2(a)_(n+1))= __

59 lim_(n to infty ) sum_(i=1)^n (n)/(n^2)+i^(2+1)=____59 $\lim_{n \to \infty }

令 Y = (1)/(n) sum_(i=1)^n X_i，则A. $\Cov(X_1, Y)= \frac{\sigma^2}{n}$.B. $Cov(X_1

,(x)_(n),(x)_(n+1) 是来自N(μ,σ^2)的样本, overrightarrow ({x)_(n)}=dfrac (1)(n)sum _(i=

设总体 X sim N(0,1)，(X_1,X_2,...,X_n) 是总体 X 的样本，令 overline(X)=(1)/(n)sum_(i=1)^nX_i

判别下列级数的绝对收敛性与收敛性：(1) sum_(n=1)^infty (i^n)/(n); (2) sum_(n=2)^infty (i^n)/(ln n

5.11 设(X1,X2,···Xn, _(n)+1) 是正态总体N(μ,σ^2)的样本, overline (X)=-|||-dfrac (1)(n)sum

1.6 总体X-N(mu,sigma^2)，x_(1),x_(2),...,x_(n)为其样本，bar(x)=(1)/(n)sum_(i=1)^nx_(i)，s