1、a(n+1)/an=(n+2)/(n+1)
a(n+1)/(n+2)=an/(n+1)
设cn=an/(n+1) 则c(n+1)=a(n+1)/(n+2),且c1=a1/(1+1)=1
即c(n+1)=cn=.=c1=1
故an/(n+1)=1
an=n+1
2、因bn=an/2^n=(n+1)/2^n
设数列｛bn｝的前n项和为Tn
Tn=b1+b2+.+bn
=2/2+3/2^2+.+(n+1)/2^n
2Tn=2/1+3/2+4/2^2+...+(n+1)/2^(n-1)
后式减前式：Tn=2+1/2+1/2^2+.+1/2^(n-1)-(n+1)/2^n
=2+(1/2)[1-(1/2)^(n-1)]/(1-1/2)-(n+1)/2^n
=3-1/2^(n-1)-(n+1)/2^n
=3-(n+3)/2^n