Question

设数列{an}前n项和为Sn,满足an=3/4Sn+1/2(n∈N*)(1)求数列{an}的通项公式；（2）令bn=nan，求数列{bn}的前n

设数列{an}前n项和为Sn,满足an=3/4Sn+1/2(n∈N*)(1)求数列{an}的通项公式；（2）令bn=nan，求数列{bn}的前n项和Tn;(3)若不等式Tn+a/n·2^2n+1-2/9＞0对任意的n∈N*恒成立，求实数a的取值范围。

Answer 1

(1)
an=(3/4)Sn+1/2 (1)
put n=1
a1= (3/4)a1+1/2
a1=2

a(n-1) =(3/4)S(n-1)+1/2 (2)
(1)-(2)
(3/4)an=an-a(n-1)
an/a(n-1)= 4
an/a1 = 4^(n-1)
an = a1. 4^(n-1)
=2^(2n-1)
(2)
bn=nan
Tn = b1+b2+..+bn
= summation n(2^(2n-1))
= (1/2)summation (n.2^2n )
consider
1+x+x^2+..+x^n = (x^(n+1)-1)/(x-1)
1+2x+...+n(x^(n-1))
= [(x^(n+1)-1)/(x-1)]'
= { (x-1)(n+1)x^n - (x^(n+1)-1) }/(x-1)^2
= { nx^(n+1)-(n+1)x^n+1 } /(x-1)^2
multiply both side by x
x+2x^2+...+nx^n = x{ nx^(n+1)-(n+1)x^n+1 } /(x-1)^2
put x = 2^2
2^2+ 2(2^4)+...+n(2^2n) = (4/9) [n2^(2n+2)-(n+1)2^(2n)+1 ]
Tn = (1/2)summation (n.2^2n )
= (1/2){(4/9) [n2^(2n+2)-(n+1)2^(2n)+1 ] }
= (2/9) [n2^(2n+2)-(n+1)2^(2n)+1 ]
(3)
Tn+a/(n·2^2(n+1))-2/9 >0
(2/9) [n2^(2n+2)-(n+1)2^(2n)+1 ] + a/(n·2^2(n+1)) - 2/9 >0
a/(n·2^2(n+1)) > (2/9) [(n+1)2^(2n) - n2^(2n+2)]
= (1/9)[2^(2n+1)](n+1 - 4n)
= (1/9)(1-3n).2^(2n+1)
a > (1/9)(1-3n)/n

Answer 2

1> 1，n=1，则a1=3/4a1+1/2 得a1=2
2，an=3/4Sn+1/2 an-1=3/4sn-1+1/2
an-an-1=3/4Sn+1/2-3/4sn-1-1/2
an-an-1=3/4an
1/4an=an-1即an=4an-1
数列｛an｝为首项为2，公比为4的等比数列。