Question

高中数学:已知数列an的前n项和为sn,且满足sn=4/3(an–1)

Answer 1

(1)
n=1时，a1=(4/3)(a1-1)
a1=4
n≥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)=2n
1/[(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)<½
Tn<½

Answer 2

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