已知数列{an}的前n项和Sn=-an-1/2^n-1+2(n为整数)(1)令bn=2^an,求证数列{bn}是等差数列,并求数列{an}的通项公式(2)令bn=n+1/n•an,求Tn=C1+C2+……+Cn

来源:学生作业帮助网 编辑:作业帮 时间:2024/05/09 04:44:33
已知数列{an}的前n项和Sn=-an-1/2^n-1+2(n为整数)(1)令bn=2^an,求证数列{bn}是等差数列,并求数列{an}的通项公式(2)令bn=n+1/n•an,求Tn=C1+C2+……+Cn

已知数列{an}的前n项和Sn=-an-1/2^n-1+2(n为整数)(1)令bn=2^an,求证数列{bn}是等差数列,并求数列{an}的通项公式(2)令bn=n+1/n•an,求Tn=C1+C2+……+Cn
已知数列{an}的前n项和Sn=-an-1/2^n-1+2(n为整数)
(1)令bn=2^an,求证数列{bn}是等差数列,并求数列{an}的通项公式
(2)令bn=n+1/n•an,求Tn=C1+C2+……+Cn

已知数列{an}的前n项和Sn=-an-1/2^n-1+2(n为整数)(1)令bn=2^an,求证数列{bn}是等差数列,并求数列{an}的通项公式(2)令bn=n+1/n•an,求Tn=C1+C2+……+Cn
用把bn、bn-1代入Sn-Sn-1=an中.