a(n+1)=6n+1 bn=3/(6n-5)(6n+1) =1/2*6/(6n-5)(6n+1) =1/2*[(6n+
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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)(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?
(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?
裂项相消法
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