高中数学:已知数列an的前n项和为sn,且满足sn=4/3(an–1)

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(1)n=1时，a1=(4/3)(a1-1)a1=4n≥2时，an=Sn-S(n-1)=(4/3)(an-1)-(4/3)[a(n-1)-1]an/a(n-1)=4，为定值。数列{an}是以4为首项，4为公比的等比数列an=4·4ⁿ⁻¹=4ⁿ数列{an}的通项公式为an=4ⁿ(2)bn=log2(an)=log2(4ⁿ)=2n1/[(bn-1)(bn+1)]=1/[(2n-1)(2n+1)]=½[1/(2n-1)- 1/(2n+1)]Tn=½[1/1 -1/3 +1/3 -1/5+...+1/(2n-1) -1/(2n+1)]=½[1- 1/(2n+1)]=½- 1/(4n+2)1/(4n+2)>0，½- 1/(4n+2)<½Tn<½