=m-1.
【答案】m-1
4.设数列{an}的前n项和为Sn,4Sn=an+2an-3,且a1,a2,a3,a4,a5成等比数列,2
当n≥5时,an>0.
(1)求证:当n≥5时,{an}成等差数列; (2)求{an}的前n项和Sn.
【解析】(1)由4S2
2n=an+2an-3,4Sn+1=an+1+2an+1-3, 得4a2
2
n+1=an+1-an+2an+1-2an, 即(an+1+an)(an+1-an-2)=0. 当n≥5时,an>0,所以an+1-an=2, 所以当n≥5时,{an}成等差数列.
(2)由4a1=a21+2a1-3,得a1=3或a1=-1, 又a1,a2,a3,a4,a5成等比数列, 而a5>0,所以a1>0,从而a1=3, 所以an+1+an=0(n≤5),q=-1,
??3(-n-1
所以a1),1≤n≤4,
n=???
2n-7,n≥5,
?所以S?32[1-(-1)n],1≤n≤4,
n=?
??n2-6n+8,n≥5.
13
备课札记 14
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