数列an的前n项和Sn且满足Sn=4/3an-1/3*2^n+1+2/3,1)求数列的首项a1与通项Sn(2)设数列bn=2^n/_数学解答

数列an的前n项和Sn且满足Sn=4/3an-1/3*2^n+1+2/3
,1)求数列的首项a1与通项Sn(2)设数列bn=2^n/Sn的前项和为Tn求证Tn<3/2

人气：360 ℃　时间：2020-03-22 01:31:08

解答

(1)n=1时,a1=S1=(4/3)a1-(1/3)·2²+2/3解得a1=2n≥2时,an=Sn-S(n-1)=(4/3)an -(1/3)·2^(n+1) +2/3-[(4/3)a(n-1)-(1/3)·2ⁿ+2/3]整理,得an=4a(n-1)+2ⁿan+2ⁿ=4a(n-1)+2^(n+1)=4a(n-1)+4·2^(n...