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1 . 在数列eqId054d5a7de5684002909fe944c86227c2中,其前eqId5f904d1376ac4de7a7d6cfea92ff6981的和是eqIddb5481de79c946c0a760143297d5eade,下面正确的是(   )
A.若eqId9cfea2d311da45488a84ad01481210f8eqIdc2d3271b09324d98954c33293515c0e9,则eqId213dac5ba2a041fc85141db50f2a8618
B.若eqIde318b67ea5d44e6894194ab3b35a6c24 ,则eqId55cb1fab71194e9ab392563fa46c6981
C.若eqIdb21fdb929d8a4de8acb013d7f44f555d ,则eqIda759ab12ab354accb63dc91a9964bf68
D.若eqId681ee32b0980433e94ff5b7e029a5760 ,且eqId76ac64e73787434b95594fdae5ed7458,则eqId5ba76b3252844111b35d8cc4c7b7aa13
2 . 已知数列eqId93e38ecd74a24cb59da79181b95bfd3a的前n项和为eqIddb5481de79c946c0a760143297d5eade,且eqIdadfbed3526604a9d9798244b57160127.
(1)求数列eqId93e38ecd74a24cb59da79181b95bfd3a的通项公式;
(2)令eqIddc3800b8c8e445fcb18049842c8122e1,求数列eqIdeeb4ca98872f4e6d91cea28f43fc0b63的前n项和eqId92ed363a54bd4a8a93df9463bb3af1f5.
解答题 | 一般(0.65) | 2021·镇远县文德民族中学校高一月考
解题方法
3 . 已知递增的等差数列eqId054d5a7de5684002909fe944c86227c2的首项是1,eqIddb5481de79c946c0a760143297d5eade是其前eqId5f904d1376ac4de7a7d6cfea92ff6981项和,且eqIdcaabbf7366f040c1b15ac1666c263e10.
(1)求eqId054d5a7de5684002909fe944c86227c2的通项公式;
(2)设eqIda334a4acd04446fba57ff32d76ccb837,求数列eqId69fa66742e6b480eb0e55f574b8b7749的前eqId5f904d1376ac4de7a7d6cfea92ff6981项和eqId92ed363a54bd4a8a93df9463bb3af1f5.
4 . 已知数列eqId93e38ecd74a24cb59da79181b95bfd3a的前eqId5f904d1376ac4de7a7d6cfea92ff6981项和为eqIddb5481de79c946c0a760143297d5eade,且eqId5c1c12d4b3ba4aaab6267324ae66c81eeqIddb5481de79c946c0a760143297d5eadeeqIdc97d581e55d345109906f78cb0654488成等差数列.
(1)求数列eqId93e38ecd74a24cb59da79181b95bfd3a的通项公式;
(2)求数列eqIda0c016c0f57441d585c912123a47eaed的前eqId5f904d1376ac4de7a7d6cfea92ff6981项和eqId92ed363a54bd4a8a93df9463bb3af1f5.
5 . 设数列eqId93e38ecd74a24cb59da79181b95bfd3a满足eqIddcae5aeb54e344b2839fda8718a2bed3
(1)求数列eqId93e38ecd74a24cb59da79181b95bfd3a的通项公式;
(2)令eqId29a70ef23c0f4a66977af8e03f7c280f,求数列eqIdeeb4ca98872f4e6d91cea28f43fc0b63的前n项和eqIddb5481de79c946c0a760143297d5eade
6 . 已知数列eqId93e38ecd74a24cb59da79181b95bfd3a中,eqIdbc36a49029fc4646bd9940dd7e4764ed,其前eqId5f904d1376ac4de7a7d6cfea92ff6981项和eqIddb5481de79c946c0a760143297d5eade满足eqIdbcae7f9c1e7a4624975a1fd02a5438bc,其中eqId10a29cfa29b242d2bf11f16d29380d81,eqId532dc4d959024b79b546428f5856e5aa
(1)求证:数列eqId93e38ecd74a24cb59da79181b95bfd3a为等差数列,并求其通项公式.
(2)设eqId4a93f30119f549ad927ba8b72a28067e,eqId92ed363a54bd4a8a93df9463bb3af1f5为数列eqIdeeb4ca98872f4e6d91cea28f43fc0b63的前n项和,求eqId92ed363a54bd4a8a93df9463bb3af1f5的最小值.
7 . 已知数列eqId93e38ecd74a24cb59da79181b95bfd3aeqIdeeb4ca98872f4e6d91cea28f43fc0b63满足:eqIdf9dc49c0cc8c4adabc02f167e11cda48eqIdc2d3271b09324d98954c33293515c0e9eqId9389483044d9400ba94b63e0f7b0fb86
(1)求数列eqId93e38ecd74a24cb59da79181b95bfd3aeqIdeeb4ca98872f4e6d91cea28f43fc0b63的通项公式;
(2)数列eqId2f151408c8a146759d59db6f8ebd0e50满足:eqId9b43ef667c4e4bcd818363e6da9c8595,其中eqId6c2146ec6df54fc9af9b2aef04489c8b,若数列eqId2f151408c8a146759d59db6f8ebd0e50的前eqId5f904d1376ac4de7a7d6cfea92ff6981项和为eqId4a772cc12bf54e07af16c18dba1a5d09,求eqId4a772cc12bf54e07af16c18dba1a5d09
解答题 | 一般(0.65) | 2021·全国高三月考(文)
8 . 已知正项数列eqId93e38ecd74a24cb59da79181b95bfd3a的前eqId5f904d1376ac4de7a7d6cfea92ff6981项和为eqIddb5481de79c946c0a760143297d5eade,且满足eqId6be932a5bf5f4bb8bef8452d83d1842beqIddb5481de79c946c0a760143297d5eadeeqId4224c303ef854035b455fe10223f984f成等差数列.
(1)求数列eqId93e38ecd74a24cb59da79181b95bfd3a的通项公式;
(2)请从以下三个条件中任意选择一个,求数列eqIdeeb4ca98872f4e6d91cea28f43fc0b63的前n项和Tn,.
条件Ⅰ:设数列eqIdeeb4ca98872f4e6d91cea28f43fc0b63满足eqId308fff60406a428d8d69ffcdb61a4607;条件Ⅱ:设数列eqIdeeb4ca98872f4e6d91cea28f43fc0b63满足eqIdf2dc287de1414354afce055c1021f825;条件Ⅲ:设数列eqIdeeb4ca98872f4e6d91cea28f43fc0b63满足eqId60613822a53b4012a08aa5b05da49018.
解答题 | 一般(0.65) | 2021·东北育才学校
解题方法
9 . 已知各项均不相等的等差数列eqId2c1230a9ef5a4968be599681d676dd2b的前4项和为10,且eqIdb5a0ee5591bb4364a22da1c2218e7bfceqId2c08dbf3fb4c466caaa6d6230082794ceqId310274b5ffe34ad58a069683d0dac787是等比数列eqIda0db2906046940e19de8782f94f34137的前3项.
(1)求eqId4224c303ef854035b455fe10223f984feqId5523abd87b714246ba23423672c79ffc
(3)设eqId88689d8c369e440cbc4f69290e69ec88,求eqId50623a2ce8b5456c9d012c2d2d27c438的前eqId5f904d1376ac4de7a7d6cfea92ff6981项和为eqIddb5481de79c946c0a760143297d5eade
10 . 已知数列eqId93e38ecd74a24cb59da79181b95bfd3a满足eqIdc2d3271b09324d98954c33293515c0e9eqIdfe4b0976e5b14c99884c2f2acc212b01
(1)设eqId93a848ead1e04982959cb7a8fc4e17d2,证明:数列eqIdeeb4ca98872f4e6d91cea28f43fc0b63是等差数列;
(2)记eqIddb5481de79c946c0a760143297d5eade为等差数列eqId93e38ecd74a24cb59da79181b95bfd3a的前eqId5f904d1376ac4de7a7d6cfea92ff6981项和,若对任意的eqId532dc4d959024b79b546428f5856e5aa,不等式eqId75ffe7ad454345418d4e0c2262f2472a恒成立,求实数eqId4d2187284c5d4de29906363f7d21f60f的最大值.