已知数列{an}的各项均为正数,前n项和为Sn,且满足2Sn=an^2+n-4求{an}的通项公式及
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![已知数列{an}的各项均为正数,前n项和为Sn,且满足2Sn=an^2+n-4求{an}的通项公式及](/uploads/image/z/1999107-27-7.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%90%84%E9%A1%B9%E5%9D%87%E4%B8%BA%E6%AD%A3%E6%95%B0%2C%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2C%E4%B8%94%E6%BB%A1%E8%B6%B32Sn%3Dan%5E2%2Bn-4%E6%B1%82%7Ban%7D%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F%E5%8F%8A)
已知数列{an}的各项均为正数,前n项和为Sn,且满足2Sn=an^2+n-4求{an}的通项公式及
已知数列{an}的各项均为正数,前n项和为Sn,且满足2Sn=an^2+n-4求{an}的通项公式及
已知数列{an}的各项均为正数,前n项和为Sn,且满足2Sn=an^2+n-4求{an}的通项公式及
用基本概念解
当n=1时,解得a1=3
(1)2Sn=an^2+n-4
(2)2(Sn-1)=(an-1)^2+n-1-4
用(1)式减去(2)式,得到
2(Sn-(Sn-1))=an^2-(an-1)^2+1
等号左边=2an,所以化简上式:
2an=an^2-(an-1)^2+1
移项,
an^2-2an-(an-1)^2+1=0
an^2-2an-[(an-1)^2-1]=0
an^2-2an-[((an-1)+1)((an-1)-1)]=0
(an-(an-1)-1)(an+(an-1)-1)=0
所以,
an-(an-1)-1=0或者an+(an-1)-1=0
因为an各项都是整数,所以an+(an-1)-1不成立,
所以an-(an-1)-1=0,即an-(an-1)=1,这是个等差数列,a1=3,公差是1
所以通项公式是:an=n+2