a(n+1)=6n+1 bn=3/(6n-5)(6n+1) =1/26/(6n-5)(6n+1) =1/2[(6n+

来源：学生作业帮编辑：大师作文网作业帮分类：数学作业时间：2024/10/07 08:28:16

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?

裂项相消法

Sn=n(n+2)(n+4)的分项等于1/6[n(n+2)(n+4)(n+5)-(n-1)n(n+2)(n+4)]吗? 1\n(n+3)+1\(n+3)(n+6)+1\(n+6)(n+9)=1\2 n+18 n为正整数,求n的值 [3n(n+1)+n(n+1)(2n+1)]/6+n(n+2)化简 3n²-n=1 求6n³+7n²-5n+2014 （数列）A(n)=（n+2）/2^n;B(n)=(6n+11)/5(n+1)试比较A(n)与B(n)大小（n∈N*）不好 Dim n% n = 1 Do Until n > 6 Print n； n = n + 2.4 Loop M=(N-1)×1+(N-2)×2+(N-3)×4+(N-4)×8+(N-5)×16+(N-6)×32+(N-7)×64 数学归纳法证明：1*n+2(n-1)+3(n-2)+…+(n-1)*2+n*1=(1/6)n(n+1)(n+2) 用数学归纳法证明：1*n+2(n-1)+3(n-2)+…+(n-1)*2+n*1=(1/6)n(n+1)(n+2) 若3n^2-n=1,求6n^3+7n^2-5n+2003的值请教初一的数学题急求证：N=52*32n+1*2n－3n*3n*6n+2能被13整除．2 2n+1 n n n n+2分已知数列an满足a1=5/6,a(n+1)=1/3an+(1/2)^(n+1),n属于N*,数列bn满足bn=a（n+1