(1)由a1+d=2和5a1+10d=15解得a1=1,d=1,∴an=n(2)由bn=n/2^n得,Tn=½+2/2^2+3/2^3+.+n/2^n,Tn/2=1/2^2+2/2^3+3/2^4+.+(n-1)/2^n+n/2^(n+1)两式相减得Tn/2=1/2+1/2^2+1/2^3+.+1/2^n-n/2^(n+1)=(1-1/2^n)-n/2^(n+1)...