你是yy中学的?.
答案:
(1)
n=1时,a1=S1=3,
n≥2时,2^(n-1)*an=Sn-S(n-1)=-6,an=-12/2^n
故an=3...(n=1时)
-12/2^n.(n≥2时)
(2)b1=3,
n≥2时,bn=n(3-log2(12/3*2^n)
=n(n+1)
即1/bn的前n项和为Tn
则T1=1/b1=1/3
n≥2时,Tn=1/b1+...+1/bn
=1/3+1/(2*3)+...+1/[n(n+1)]
=1/3+(1/2-1/3)+...+[1/n-1/(n+1)]
=5/6-1/(n+1),n=1时也满足
故Tn=5/6-1/(n+1)