已知等比数列{an}的前n项和为Sn,且an是Sn与2的等差中项,等差数列{bn}中,b1=2,点P(bn,bn+1)在直线y=x+2上求a1和a2的值求数列{an}和{bn}的通项an和bn设Cn=an*bn,求数列{cn}的前n项和TnPs:答出第3问才有分拿,答对前2问没有分针对一楼的回答好象有点点错误...Tn=1*2^2+2*2^3+3*2^4+……+n*2^(n+1) →2Tn= 1*2^3+2*2^4+……+(n-1)*2^(n+1)+n*2^(n+2) 两式相减 -Tn=Tn-2Tn=1*2^2+1*2^3+1*2^4+……+1*2^(n+1)-n*2^(n+2) 麻烦检查一下有无错误

问题1:2a1=S1+2=a1+2
→a1=2
2a2=S2+2=a1+a2+2
→a2=4
问题2:an是Sn与2的等差中项
→2an=Sn+2
2a(n-1)=S(n-1)+2
两式相减
2an-2a(n-1)=Sn-S(n-1)=an
→an=2a(n-1)
→an=2a(n-1)=2^2a(n-2)=……＝2^(n-1)a1=2^n
点P(bn,bn+1)在直线y=x+2上
→b(n+1)=bn+2
→bn=b(n-1)+2=b(n-2)+2*2=b(n-3)+2*3=……=b1+2(n-1)=2+2(n-1)=2n
问题3:设Cn=an*bn=n*2^(n+1)
Tn=1*2^2+2*2^3+3*2^4+……+n*2^(n+1)
→2Tn= 1*2^3+2*2^4+……+(n-1)*2^(n+1)+n*2^(n+2)
两式相减
-Tn=Tn-2Tn=1*2^2+1*2^3+1*2^4+……+1*2^(n+1)-n*2^(n+2)
=4+8+16+……+2^(n+1)-n*2^(n+2)
=4+4+8+16+……+2^(n+1)-n*2^(n+2)-4
=8+8+16+……+2^(n+1)-n*2^(n+2)-4
=……
=2^(n+2)-n*2^(n+2)-4
=-(n-1)*2^(n+2)-4
故Tn=(n-1)*2^(n+2)+4