满足
dk?1?4为常数,所以数列?dk?为等比数列 dk(3)①当k为奇数时,
1k?15?Ck25k?2???(?1)k4k(5?1)k5k?Ckdk???555,
11k?2?5k?1?Ck5?Ck25k?3???Ckk?150(?1)k?1?54k?1(5?1)k?111k?12k?2k0k??5k?Ck5?C5???C5(?1)?同样,可得dk?1?, ?1k?1k?1555所以,集合xdk?x?dk?1,x?Z的元素个数为(dk?1?)?(dk?)?1
??151533(4k?1)?dk?1?dk??;
553?(4k?1)②当k为偶数时,同理可得集合?xdk?x?dk?1,x?Z?的元素个数为
5错误！未找到引用源。本题共有3小题,第1小题满分4分,第2小题满分6分,第3小题满
分8分.
解:(1)an?1?Sn?3n?Sn?1?2Sn?3n,bn?Sn?3n,n?N?,当a?3时,
bn?1Sn?1?3n?12Sn?3n?3n?1=2,所以?bn?为等比数列. ??bnSn?3nSn?3nb1?S1?3?a?3,bn?(a?3)?2n?1.
(2) 由(1)可得Sn?3n?(a?3)?2n?1
an?Sn?Sn?1,n?2,n?N?
an?1?; an??n?1n?2n?2?2?3?(a?3)?2?a?a1 ,a??9 an?1?an,?2?an?1?ann?2所以a??9,且a?3.所以a的最小值为 (3)由(1)当a?4时,bn?2n?1
n当n?2时,Cn?3?2?4???2n?2?1,C1?3,
所以对正整数n都有Cn?2n?1.
由tp?2n?1,tp?1?2n,(t,p?N?且t?1,p?1),t只能是不小于3的奇数.
①当p为偶数时,t?1?(t因为tp2p2pp2?1)(t?1)?2n,
p2?1和t?1都是大于1的正整数,
p2所以存在正整数g,h,使得t?1?2,t?1?2h,
gp22g?2h?2,2h(2g?h?1)?2,所以2h?2且2g?h?1?1?h?1,g?2,相应的
n?3,即有C3?32,C3为“指数型和”;
2p?1p2p?1②当p为奇数时,t?1?(t?1)(1?t?t???t),由于1?t?t???t是p个奇数之和,仍为奇数,又t?1为正偶数,所以(t?1)(1?t?t2???tp?1)?2n 不成立,此时没有“指数型和”.
错误！未找到引用源。解:由题意知a1?2,且
ban?2n??b?1?Sn ban?1?2n?1??b?1?Sn?1
两式相减得b?an?1?an??2??b?1?an?1
n即an?1?ban?2n ① (1)当b?2时,由①知an?1?2an?2n 于是an?1??n?1??2?2an?2??n?1??2
nnn?2?an?n?2n?1?
n?1又a1?1?2n?1?1?0,所以an?n?2是首项为1,公比为2的等比数列.
??故知,bn?2n?1,
n?1再由bn?an?n?2另解:
,得an??n?1?2n?1.
an?1an1?? 2n?12n2a11?a??1是首项为,公差为的等差数列, ??n1n?22?2??ann?1n?1?1?? 2n22?an??n?1??2n?1 bn??n?1??2n?1?n?2n?1?2n?1
(2)当b?2时,由①得
an?1?111???2n?1?ban?2n??2n?1?b?an??2n? 2?b2?b2?b??若b?0,Sn?2n 若b?1,an?2n,Sn?2n?1?2
1,数列?an?若b?0、??2(1?b)1?为首项,以b为公比的等比数列,故 ?2n?是以
2?b2?b?an?12(1?b)n?1?2n??b, 2?b2?b1an?2n??2?2b?bn?1
2?b12(1?b)Sn?2?22?23?????2n?1?b1?b2?????bn?1
2?b2?b??????2(2n?bn)Sn?
2?bb?1时,Sn?2n?1?2符合上式
2(2n?bn)所以,当b?0时,Sn?
2?bn当b?0时,Sn?2
另解:
当n?1时,S1?a1?2 当n?2时,?ban?2??b?1?Sn
n
?b?Sn?Sn?1??2n??b?1?Sn
?Sn?bSn?1?2n
若b?0,Sn?2n 若b?0,两边同除以2得
nSnbSn?1???1 2n22n?1令
SnSnbSn?1bSn?12?2m?m???1?m?m??(n?1?) ,即nn?1n222222b2?2m2得m? bb?2由m??{Snb2b?}是以为首项,为公比的等比数列 n22b?2b?2?Sn2bbn?1???(), 2nb?2b?222(2n?bn)所以,当b?0时,Sn?
2?b
错误！未找到引用源。