设等比数列an的前n项和为sn,点(n,sn)在fx=3x²-2x的图像上,①求an通项公式②设bn=3/an*a（n+1）,Tn是其前n项和,求使得|Tn-1/2|＜1/100的最小整数n

Question

设等比数列an的前n项和为sn,点(n,sn)在fx=3x²-2x的图像上,①求an通项公式 ②设bn=3/an*a（n+1）,
Tn是其前n项和,求使得|Tn-1/2|＜1/100的最小整数n

Answer 1

(1)
x=n f(x)=Sn代入函数方程
Sn=3n²-2n
n=1时,a1=S1=3×1²-2×1=1
n≥2时,
an=Sn-S(n-1)=3n²-2n-[3(n-1)²-2(n-1)]=6n-5
n=1时,a1=6×1-5=1,同样满足通项公式
数列{an}的通项公式为an=6n-5
(2)
bn=3/[an·a(n+1)]=3[(6n-5)(6n+1)]=(1/2)[1/(6n-5)-1/(6n+1)]
Tn=b1+b2+...+bn
=(1/2)[1/1-1/7+1/7-1/13+...+1/(6n-5)-1/(6n+1)]
=(1/2)[(1/1+1/7+...+1/(6n-5))-(1/7+1/13+...+1/(6n+1))]
=(1/2)[1-1/(6n+1)]
=3n/(6n+1)
|Tn-1/2|