已知数列{an}满足a1=1,an>0,sn是数列{an}的前n项和,对任意n是正数,有2sn=p(2an^2+an-1)(P为常数)(1)求p的值,(2)计算a2,a3的值,并求数列{an}的通项公式.
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已知数列{an}满足a1=1,an>0,sn是数列{an}的前n项和,对任意n是正数,有2sn=p(2an^2+an-1)(P为常数)(1)求p的值,(2)计算a2,a3的值,并求数列{an}的通项公式.
已知数列{an}满足a1=1,an>0,sn是数列{an}的前n项和,对任意n是正数,有2sn=p(2an^2+an-1)(P为常数)
(1)求p的值,(2)计算a2,a3的值,并求数列{an}的通项公式.
已知数列{an}满足a1=1,an>0,sn是数列{an}的前n项和,对任意n是正数,有2sn=p(2an^2+an-1)(P为常数)(1)求p的值,(2)计算a2,a3的值,并求数列{an}的通项公式.
n=1时,2S1=2a1=p(2a1^2+a1-1),所以p=1
故 2Sn=2an^2+an-1
n=2时,2S2=2(a1+a2)=2a2^2+a2-1 ,故 a2=3/2
同理 a3=2,
.
归纳有,an=(n+1)/2
用数学归纳法证明即可
(法二,求an的通项
2Sn=2an2+an-1
2Sn-1=2an-1^2+an-1 -1
上式减下式,有 2an=2(an^2-an-1^2)+(an-an-1)
整理:an+an-1=2(an^2-an-1^2)=2(an+an-1)(an-an-1)
因为:an+an-1=0时,不合题意;
即an-an-1=1/2=d
公差为1/2 ,首项为1,故an=1+(n-1)/2=(n+1)/2
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