已知各项均为正数的数列{an}的前n项和为Sn,满足S1>1,且6Sn=(an+1)×(an+2),求通项公式
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![已知各项均为正数的数列{an}的前n项和为Sn,满足S1>1,且6Sn=(an+1)×(an+2),求通项公式](/uploads/image/z/5455683-27-3.jpg?t=%E5%B7%B2%E7%9F%A5%E5%90%84%E9%A1%B9%E5%9D%87%E4%B8%BA%E6%AD%A3%E6%95%B0%E7%9A%84%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2C%E6%BB%A1%E8%B6%B3S1%EF%BC%9E1%2C%E4%B8%946Sn%3D%EF%BC%88an%2B1%29%C3%97%28an%2B2%29%2C%E6%B1%82%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F)
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已知各项均为正数的数列{an}的前n项和为Sn,满足S1>1,且6Sn=(an+1)×(an+2),求通项公式
已知各项均为正数的数列{an}的前n项和为Sn,满足S1>1,且6Sn=(an+1)×(an+2),求通项公式
已知各项均为正数的数列{an}的前n项和为Sn,满足S1>1,且6Sn=(an+1)×(an+2),求通项公式
当n=1时.6a1=(a1+1)(a1+2),a1=2或1(因为S1gt;1,则舍)
当n用n-1写时,得:6Sn-1 =(an-1 +1)(an-1 +2),与6Sn=(an+1)(an+2)做差,得:
6an=(an+1)(an+2)-(an-1 +1)(an-1 +2)自己化(移项,分解因式)即得
an-an-1 =3,{an}是以2为首项,3为公差的等差数列.an=3n-1
6Sn=(an+1)×(an+2)=an^2+3an+2
6a1=6S1=a1^2+3a1+2
a1=1或2
又S1=a1>1
a1=2
n>=2时
6Sn=an^2+3an+2
6S(n-1)=a(n-1)^2+3a(n-1)+2
an^2-a(n-1)^2-3an-3a(n-1)=0
(an+a(n-1))(an-a(n-1)-3)=0
又an为正项数列
an+a(n-1)>0
an-a(n-1)-3=0
a1=2
an=3n-1