+dfrac (sin npi )(n+dfrac {1)(n)}]

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

+dfrac (1)({n)^2+npi })=1；；

+dfrac (sin n)({2)^n} ;-|||-(2) _(n)=1+dfrac (1)({2)^2}+dfrac (1)({3)^2}+... +df

+sqrt (1+cos dfrac {npi )(n)}] -|||-= __

+dfrac (1)({n)^2+npi })= 1..

一、选择题-|||-3.设 =(sin )^2x, 则 ^(n+1)=() .-|||-(A) sin (2x+dfrac (npi )(2)) (B) ^ns

__-|||-lim _(narrow infty )([ sin (dfrac {pi )(4)+dfrac (1)(n))] }^n=( )A.

24.已知 ((m)^2+9)((n)^2+1)=5-10mn ,求:-|||-(1) (n+dfrac (1)(n));-|||-(2) ((m-dfrac

若数列_(n)=sin dfrac (n)(2)pi , 则数列_(n)=sin dfrac (n)(2)pi 发散正确错误若数列,则数列发散正确错误

+dfrac (1)(n))}^dfrac (1{n)};