-1 0 -1 0 0 的值为 ()-|||-A 1-|||-B ((-1))^dfrac (n(n-1){2)}-|||-C -1-|||-D ((-1))^dfrac (n(n+1){2)}

的值为 .-|||-n-1 0 ...0 0 0-|||-0 0 ...0 0 n-|||-A ((-1))^dfrac ((n-1)(n-2){2)n!}!-

1-|||-1 a1 0 ····0-|||-计算 n+1 阶行列式 _(n+1)= 1 0 a2 ... 0 _(i)neq 0(i=1,2,... ,n).

+(n)^3);-|||-(2) lim _(narrow infty )n[ dfrac (1)({(n+1))^2}+dfrac (1)({(n+2))^2

n-1 n-|||-1 2 ... n-1 0-|||-::-|||-1 2 ... 0 0-|||- ... 0 0;;

下列级数绝对收敛的是()A.sum _(n=1)^infty dfrac ({(-1))^n+1}(2n+1)B.sum _(n=1)^infty dfrac

+dfrac (1)(n(n+1)) =-|||-(3) lim _(narrow infty )(dfrac (1)(2)+dfrac (3)({2)^2}+

(6)收敛, lim _(narrow infty )dfrac ({2)^n-1}({3)^n}=0.-|||-(7) n-dfrac {1)(n)} 发

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

[ dfrac (sin z)({z)^2},0] =-|||-A 1-|||-B .-1-|||-C dfrac (1)(2)

_(n)=((-1))^n+1dfrac (1)(sqrt {n)}-|||-C. _(n)=sin dfrac (npi )(2)-|||-D. _(n)=d