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
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![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](/uploads/image/z/2981909-29-9.jpg?t=%E5%B7%B2%E7%9F%A5%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2C%E4%B8%94an%E6%98%AFSn%E4%B8%8E2%E7%9A%84%E7%AD%89%E5%B7%AE%E4%B8%AD%E9%A1%B9%2C%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%7Bbn%7D%E4%B8%AD%2Cb1%3D2%2C%E7%82%B9P%28bn%2Cbn%2B1%29%E5%9C%A8%E7%9B%B4%E7%BA%BFy%3Dx%2B2%E4%B8%8A%E6%B1%82a1%E5%92%8Ca2%E7%9A%84%E5%80%BC%E6%B1%82%E6%95%B0%E5%88%97%7Ban%7D%E5%92%8C%7Bbn%7D%E7%9A%84%E9%80%9A%E9%A1%B9an%E5%92%8Cbn%E8%AE%BECn%3Dan%2Abn%2C%E6%B1%82%E6%95%B0%E5%88%97%7Bcn%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8CTnPs%3A%E7%AD%94%E5%87%BA%E7%AC%AC3%E9%97%AE%E6%89%8D%E6%9C%89%E5%88%86%E6%8B%BF%2C%E7%AD%94)
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
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Tn
Ps: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
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
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