(I)根据框图可知an+1=an+2整理得an+1-an=2,根据等差数列的定义判断出{an}为等差数列,进而根据等差数列的通项公式求得an,根据bn+1=3bn,整理得判断出{bn}为等比数列,根据首项和公比求得{bn}的通项公式.
(II)根据(1)中求得的an和bn,求得cn,进而利用错位相减法求得答案.
【解析】
(I)依框图得,an+1=an+2,a1=1,
即an+1-an=2,∴数列{an}是首项为1,公差为2的等差数列
∴an=1+(n-1)×2=2n-1
又bn+1=3bn,b1=3,
即,∴数列{bn}是首项为3,公比为3的等比数列
∴bn=3×3n-1=3n
(II)由(I)得cn=anbn=(2n-1)•3n
∵数列{cn}的前n和为Tn∴Tn=c1+c2+c3++cn-1+cnTn=1×31+3×32+5×33++(2n-3)×3n-1+(2n-1)×3n①
∴3Tn=1×32+3×33+5×34++(2n-3)×3n+(2n-1)×3n+1②
将①-②得:-2Tn=3+2×32+2×33+2×34++2×3n-(2n-1)×3n+1=-3+2(3+32+33+34++3n)-(2n-1)×3n+1
=
Tn=(n-1)×3n+1+3