；
；
+dfrac (1)({n)^2+npi })= 1..
_(n)=((-1))^n+1dfrac (1)(sqrt {n)}-|||-C. _(n)=sin dfrac (npi )(2)-|||-D. _(n)=d
+dfrac (sin npi )(n+dfrac {1)(n)}]
+sqrt (1+cos dfrac {npi )(n)}] -|||-= __
积分 (int )_(dfrac {1)(2)}^dfrac (1{2)}(e)^npi dz=________________.(int )_(dfrac {
一、选择题-|||-3.设 =(sin )^2x, 则 ^(n+1)=() .-|||-(A) sin (2x+dfrac (npi )(2)) (B) ^ns
+dfrac (sin n)({2)^n} ;-|||-(2) _(n)=1+dfrac (1)({2)^2}+dfrac (1)({3)^2}+... +df
+(n)^3);-|||-(2) lim _(narrow infty )n[ dfrac (1)({(n+1))^2}+dfrac (1)({(n+2))^2
+dfrac (1)(n(n+1)) =-|||-(3) lim _(narrow infty )(dfrac (1)(2)+dfrac (3)({2)^2}+
+dfrac (n)({n)^2+n+n})=dfrac (1)(2)证明：