已知数列｛an｝的前n项和Sn=—n平方+2kn（k∈N）,且Sn的最大值4,令bn=5—an/3的n次方,数列｛bn｝

Sn =-n^2+2kn
let
f(x) =-x^2+2kx
f'(x) = -2x+2k=0
x=k
f'(k)= k^2 = 4
k=2
Sn =-n^2+4n
a1= 3
an = Sn - S(n-1)
= -(2n-1) +4
= -2n+5
bn =(5- an)/3^n
= 2n/3^n
= 2(n.1/3^n)
Tn = b1+b2+...+bn
= 2S
S = 1.(1/3)^1+2.(1/3)^2+.+n(1/3)^n (1)
(1/3)S = 1.(1/3)^2+2.(1/3)^3+.+n(1/3)^(n+1) (2)
(1)-(2)
(2/3)S = (1/3 +1/3^2+...+1/3^n) -n(1/3)^(n+1)
= (1/2)(1- 1/3^n) - n(1/3)^(n+1)
= 1/2 - (1/2)(2n+3).3^(n+1)
S = 3/4 -(3/4)(2n+3).3^(n+1)
Tn = 2S
= 3/2 -(3/2)(2n+3).3^(n+1)
< 3/2