一、（20分）计算下列各题：
kk?1．求极限 lim?(1?)sin2
n??nnk?1n?1解法1
kk? 因?(1?)sin2
nnk?1n?1?12sin?2n22sin?(1?n）nk?1n?1kk?2sin?2n2
?k2k???2k???(1?）(cos?cos) 22??n2n2n2sin2k?12n1n?1?n2k2k???2k???(1?）(cos?cos) ?n?k?12n22n2n?1n?1k2k???n2n?1k2k??? ??(1?）cos?(1?）cos?22?k?1n2n?k?1n2nn2k2k???n2??(1?）cos?2?k?1n2n?n?1n2?(1?k?2nk?12k???）cosn2n2
1?n2n?1(2n?1)?n2n?112k??? ?(1?)cos2?(1?）cos?cos??n2n?n2n2?k?2n2n21?n21(2n?1)?nn?1(2k?1)?（*） ?(1?)cos2?(2?）cos?cos?22?n2n?n2n?k?22n而
1(2k?1)??cos??2n2k?22sin22nn?1n2n2?2cosk?2n?1(2k?1)??sin
2n22n2?12sin?2n2?[sink?2n?1k?(k?1)??sin] 22nn 1
(n?1)???sinn2n2 ??2sin22n?(n?2)?cossin22n2n（**） ?sin2n2将（**）代入（*），然后取极限，得
原式?lim[n??sin?n1?n1(2n?1)?n(1?)cos2?(2?）cos?2?n2n?n2n?22cos?2nsinsin?(n?2)?2n2]
2n21n21(2n?1)2?22n3?(n?2)??lim[(1?)?(2?）(1?)?cossin] n???n?n8n4?22n2n21n21?22n3?(n?2)??lim[(1?)?(2?）(1?2)?2cossin] n???n?n2n?2n2n21n21?22n3?2(n?2)?(n?2)3?3?lim[(1?)?(2?）(1?2)?2(1?2)(?)]26n???n?n2n?8n2n48n1n21?22n3?2???3?lim[(1?)?(2?）(1?2)?2(1?2)(?2?)] 3n???n?n2n?8n2nn48nn2n2n2n2
?
5? 62上式中含n的项的系数为项系数为??1?248解法2 Step 1
n?1?????????5?6?6
?2?1?0，含n的项的系数为
1??1??1?(?2)?0，常数
1k? 因?sin2??nk?12sin22n?12sin?2sink?1n?1n?1k??sin n22n2?2n2?(cosk?12k???2k????cos) 222n2n?故
12sin?2n2(cos?2n2?cos(2n?1)?) 2n22
lim?sinn??k?1n?1k?1?(2n?1)??lim(cos?cos)n2n??2sin?2n22n22n2
?limn??1?n2(cos?2n2?cos(2n?1)?) 22n?limn??2n2?sin(n?1)??sin 2n22n?2n2(n?1)????lim 2n???2n2n2Step 2因?cosk?2n(2k?1)? 22n(2k?1)??sin
2n22n2?12sin12sin?2n2?2cosk?2n??2n2?[sink?2nk?(k?1)??sin] n2n2sin?因此
(n?1)?(n?1)?n??cossin?sin2n22n2 n2n2???2sin2sin22n2nn?11kk??sin?n22sin?k?1n2n2n?1?2nsinnk?1n?1kk?2sin?2n2
k(2k?1)?n?1k(2k?1)??[?cos?cos] ?22?k?1n2nn2nk?12sin22n1nk(2k?1)?k?1(2k?1)??[?cos?cos] ?22?k?1n2nn2nk?22sin22n1n?1?1?n?1(2n?1)?n?11(2k?1)???cos?cos?cos?222? ??n2nn2nn2nk?2?2sin2?2n1
3
n?1?(2n?1)?1(2k?1)???cos?cos?cos?222?(*) ??n2n2nn2nk?2?2sin2?2n1(n?1)?(n?1)??cossin22?11?(2n?1)?2n2n?cos2?cos?2??n?2n2n2sin2?nsin22n?2n于是
kk?lim?sin2 n??nk?1nn?1??? ??(n?1)?(n?1)??cossin22?11?(2n?1)?2n2n?limcos2?cos?2n????n?2n2n2sin2?nsin22n?2n(n?1)?(n?1)???cossin22222?n?1(2n?1)?2n2n?lim??（1?）?? 4n????8n?n?2n???(n?1)2?2(n?1)?(n?1)3?3?（1?)(?)?426n2?1?28n2n48n?lim??1?2??
n????2n?n???2n????? ??n2?1?2?2n?1(n?1)3?2??lim??1?2?（1?2)(?)? 5n???n2n8nn24n??n2?1?2?2(n?1?2??lim??1?2?（1?2)(?) 2?n???2n8nn24n??nn2?1?2?21?2??lim??1?2?（1?2)(1??) 2?n???n2n8nn24n??n21?21?2?2?lim（?1?2?1???） n???n2nn24n28n2n2?2?2?2?lim（2??2） 2n???2n24n8n63??5?原式???
236
??2????
4
2．计算???axdydz?(z?a)2dxdyx?y?z222，其中 ?为下半球面
z??a2?x2?y2 的上侧, a?0.
解 记?1为平面 z?0,x2?y2?a2 的上侧，?2为下半球面 z??a2?x2?y2 的下侧，?是由?1和?2所围成的立体，则
22axdydz?(z?a)dxdy?a????dxdy??1?124adxdy??a， ??x2?y2?a2设x?rcos?,y?rsin?,则
?1??2??axdydz?(z?a)?2dxdy
????(a?0?2z?2a)dxdydz ????(2z?3a)dxdydz
??x2?y2?a2??dxdy?0?a2?x2?y2(2z?3a)dz
?x2?y2?a2??[z2?3az]0?a2?x2?y2dxdy
?x2?y2?a2222222(?a?x?y?3aa?x?y)dxdy ???0?r?a0???2?2222(?a?r?3aa?r)rdrd? ???2??(?a2?r2?3aa2?r2)rdr
0a???(?a2?r2?3aa2?r2)d(r2)
0a???(?a?u?3a(a?u))du
0a22212??u????a2u??4a(a2?u)?
2??0232a27?a4?
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