1001,S偶=a2+a4+a6+…+a2000=1000a1001.答案:16.解:原式=答案:三、解答题17.解:(1)由题意得(a1+d)(a1+13d)=(a1+4d)2(d>0)解得d=2,∴an=2n-1,可得bn=3n-1(2)当n=1时,c1=3;当n≥2时,由,得cn=2·3n-1,故故c1+c2+c3+…+c2003=3+2×3+2×32+…+2×32002=32003.18.解:(1)∵an+1=can+1-c,∴an+1-1=c(an-1),∴数列{an-1)是首项为a-1≠0,公比为c≠0的等比数列,∴an-1=(a――1,即:an=(a――1+1(2)当时,,则,利用“差比数列”的求和方法有:.19.解:依题意,(1)∵an>0,bn>0,∴an+1=bn·bn+1,同理:an=bn-1·bn(n≥2)∴2bn2=bn-1bn+bnbn+1,∴2bn=bn-1+bn+1(n≥2),∴{bn}是等差数列.(2)∵{bn}是等差数列,∴bp-q+bp+q=2bp,∴,(3)由a1=1,b1=及①②两式易得a2=3,b2=,∴{bn}中公差,∴,∴.③∴,∴,∴,20.解:(1)设{an}公差为d,由题意易知d≥0,且d∈N,则{an}通项an=3+(n-1)d,前n项和.再设{bn}公比为q,则{bn}通项bn=qn-1由b2S2=64可得q·(6+d)=64①又{bn}为公比为64的等比数列,∴,∴qd=64②联立①、②及d≥0,且d∈N可解得q=8,d=2.∴{an}通项公式an=2n+1,{bn}通项公式bn=8n-1,(2)由(1)知,n∈N*∴,n∈N*.21.(1)证明由已知S1=a1=a,Sn=aqn-1,∴Sn-1=aqn-2,∴当n≥2时,an=Sn-Sn-1=a(q-1)qn-2.∵=q,∴{an}是当n≥2时公比为q的等比数列.
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