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已知数列{an}的各项均为正数,Sn是数列{an}的前n项和,且4Sn=an2+2an-3.
(1)求数列{an}的通项公式;
(2)已知bn=2n,求Tn=a1b1+a2b2+…+anbn的值.
人气:331 ℃ 时间:2020-04-24 03:10:26
解答
(1)当n=1时,a1=s1=14a21+12a1−34,解出a1=3,又4Sn=an2+2an-3①当n≥2时4sn-1=an-12+2an-1-3②①-②4an=an2-an-12+2(an-an-1),即an2-an-12-2(an+an-1)=0,∴(an+an-1)(an-an-1-2)=0,∵an+an-1>0∴a...
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