和Sn可求数列的通项公式,通过通项公式判断{an}类型.Р解析:∵log5(Sn+1)=n,∴Sn+1=5n,∴Sn=5n-1 (n∈N+), Р∴a1=S1=51-1=4,Р当n≥2时,an=Sn-Sn-1=(5n-1)-(5n-1-1)=5n-5n-1=5n-1(5-1)=4×5n-1Р而n=1时,4×5n-1=4×51-1=4=a1,Р ∴n∈N+时,an=4×5n-1Р由上述通项公式,可知{an}为首项为4,公比为5的等比数列.Р举一反三:Р【变式1】},=2n+3n,+1-pCn}为等比数列,求常数p。Р【答案】p=2或p=3;Р∵{Cn+1-pCn}是等比数列,Р∴对任意n∈N且n≥2,+1-pCn)2=(Cn+2-pCn+1)(Cn-pCn-1)Р∵Cn=2n+3n,∴[(2n+1+3n+1)-p(2n+3n)]2=[(2n+2+3n+2)-p(2n+1+3n+1)]·[(2n+3n)-p(2n-1+3n-1)]Р即[(2-p)·2n+(3-p)·3n]2=[(2-p)·2n+1+(3-p)·3n+1]·[(2-p)·2n-1+(3-p)·3n-1]Р整理得:,解得:p=2或p=3,Р+1-pCn≠0,故p=2或p=3为所求.Р【变式2】设{an}、{bn}=an+bn,}不是等比数列.Р【证明】设数列{an}、{bn}的公比分别为p, q,且p≠qР}不是等比数列,只需证.Р∵,Р∴,Р又∵ p≠q, a1≠0, b1≠0,Р∴即Р∴}不是等比数列.Р【变式3】判断正误:Р(1){an}为等比数列a7=a3a4;Р(2)若b2=ac,则a,b,c为等比数列;Р(3){an},{bn}均为等比数列,则{anbn}为等比数列;Р(4){an}是公比为q的等比数列,则、仍为等比数列;Р(5)若a,b,c成等比,则logma,logmb,logmc成等差.Р【答案】
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