【解析】(1)当n 1时,an Sn Sn 1 k(cn cn 1) 则an Sn Sn 1 k(cn cn 1)
a6 k(c6 c5),a3 k(c3 c2)
a6c6 c5
32 c3 8,∴c=2.∵a2=4,即k(c2 c1) 4,解得k=2,∴an 2n(n)1) a3c c
当n=1时,a1 S1 2 综上所述an 2n(n N*) (2) nan n2n,则
Tn 2 2 22 3 23 n2n(1)
2Tn 1 2 2 2 3 2 (n 1)2 n2
2
3
4
n
n 1
(2)
(1)-(2)得
Tn 2 22 23 2n n2n 1 Tn 2 (n 1)2n 1