a(n+1)=6n+1 bn=3/(6n-5)(6n+1) =1/2*6/(6n-5)(6n+1) =1/2*[(6n+1)-(6n-5)]/(6n-5)(6n+1) =1/2*[1/(6n-5)-1(2006•湖北)已知二次函数y=f(x)的图象经过坐标原点,其导函数为f′(x)=6x-2,数列{an}的前n项和为Sn,点(n,Sn
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![a(n+1)=6n+1 bn=3/(6n-5)(6n+1) =1/2*6/(6n-5)(6n+1) =1/2*[(6n+1)-(6n-5)]/(6n-5)(6n+1) =1/2*[1/(6n-5)-1(2006•湖北)已知二次函数y=f(x)的图象经过坐标原点,其导函数为f′(x)=6x-2,数列{an}的前n项和为Sn,点(n,Sn](/uploads/image/z/11140050-66-0.jpg?t=a%28n%2B1%29%3D6n%2B1+bn%3D3%2F%286n-5%29%286n%2B1%29+%3D1%2F2%2A6%2F%286n-5%29%286n%2B1%29+%3D1%2F2%2A%5B%286n%2B1%29-%286n-5%29%5D%2F%286n-5%29%286n%2B1%29+%3D1%2F2%2A%5B1%2F%286n-5%29-1%EF%BC%882006%26%238226%3B%E6%B9%96%E5%8C%97%EF%BC%89%E5%B7%B2%E7%9F%A5%E4%BA%8C%E6%AC%A1%E5%87%BD%E6%95%B0y%3Df%EF%BC%88x%EF%BC%89%E7%9A%84%E5%9B%BE%E8%B1%A1%E7%BB%8F%E8%BF%87%E5%9D%90%E6%A0%87%E5%8E%9F%E7%82%B9%2C%E5%85%B6%E5%AF%BC%E5%87%BD%E6%95%B0%E4%B8%BAf%E2%80%B2%EF%BC%88x%EF%BC%89%3D6x-2%2C%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2C%E7%82%B9%EF%BC%88n%2CSn)
a(n+1)=6n+1 bn=3/(6n-5)(6n+1) =1/2*6/(6n-5)(6n+1) =1/2*[(6n+1)-(6n-5)]/(6n-5)(6n+1) =1/2*[1/(6n-5)-1(2006•湖北)已知二次函数y=f(x)的图象经过坐标原点,其导函数为f′(x)=6x-2,数列{an}的前n项和为Sn,点(n,Sn
a(n+1)=6n+1 bn=3/(6n-5)(6n+1) =1/2*6/(6n-5)(6n+1) =1/2*[(6n+1)-(6n-5)]/(6n-5)(6n+1) =1/2*[1/(6n-5)-1
(2006•湖北)已知二次函数y=f(x)的图象经过坐标原点,其导函数为f′(x)=6x-2,数列{an}的前n项和为Sn,点(n,Sn)(n∈N*)均在函数y=f(x)的图象上.
(Ⅰ)求数列{an}的通项公式;
(Ⅱ)设bn=3anan+1
,Tn是数列{bn}的前n项和,求使得Tn<m
20
对所有n∈N*都成立的最小正整数m;
bn=3*[1/(6n-5)×1/(6n+1)]
=3*(1/6)*[1/(6n-5)-1/(6n+1)]
=(1/2)*[1/(6n-5)-1/(6n+1)]
是怎样计算出来的,裂项相消法?
a(n+1)=6n+1 bn=3/(6n-5)(6n+1) =1/2*6/(6n-5)(6n+1) =1/2*[(6n+1)-(6n-5)]/(6n-5)(6n+1) =1/2*[1/(6n-5)-1(2006•湖北)已知二次函数y=f(x)的图象经过坐标原点,其导函数为f′(x)=6x-2,数列{an}的前n项和为Sn,点(n,Sn
裂项相消法