(1)a1+a2+a3=7
a4+a5+a6=56
所以q^3=8 q=2
a1=1
所以an=2^(n-1)
（2）设bn=n/an=n/2^(n-1)
Sn=1+2/2^1+3/2^2+……+n/2^(n-1)
1/2Sn=1/2^1+2/2^2+……+(n-1)/2^(n-1)+n/2^n
两式相减 1/2Sn=1+(1/2+1/2^2+1/2^3+……+1/2^(n-1)）-n/2^n
=2-1/2^(n-1)-n/2^n
Sn=4-1/2^(n-2)－n/2^(n-1)