1、a3=6,S3=12
所以a1+2d=6,3a1+3d=12
所以a1=2,d=2
所以{an}的通項公式為an=2+(n-1)*2=2n
2、Sn=2n+n(n-1)=n²+n=n(n+1)
1/Sn=1/n-1/(n+1)
Bn=1/S1+1/S2+······+1/Sn
=1-1/2+1/2-1/3+1/3-1/4+······+1/n-1/(n+1)
=1-1/(n+1)
=n/(n+1)
3、令Cn=2^(n-1)an,則Cn=2n×2^(n-1)=n×2ⁿ
Cn=C1+C2+...+Cn=1×2+2×2²+3×2³+...+n×2ⁿ
2Cn=1×2²+2×2³+...+(n-1)×2ⁿ +n×2^(n+1)
Cn-2Cn=-Cn=2+2²+...+2ⁿ-n×2^(n+1)
Cn=n×2^(n+1)-(2+2²+...+2ⁿ)
=n×2^(n+1)-2×(2ⁿ-1)/(2-1)
=n×2^(n+1)-2^(n+1) +2
=(n-1)×2^(n+1) +2