数列{an}为等差数列,数列{bn}满足bn=2an+1+a2n-1,证明{bn}为等差数列
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![数列{an}为等差数列,数列{bn}满足bn=2an+1+a2n-1,证明{bn}为等差数列](/uploads/image/z/8633051-35-1.jpg?t=%E6%95%B0%E5%88%97%EF%BD%9Ban%EF%BD%9D%E4%B8%BA%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%2C%E6%95%B0%E5%88%97%EF%BD%9Bbn%EF%BD%9D%E6%BB%A1%E8%B6%B3bn%3D2an%2B1%2Ba2n-1%2C%E8%AF%81%E6%98%8E%EF%BD%9Bbn%EF%BD%9D%E4%B8%BA%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97)
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数列{an}为等差数列,数列{bn}满足bn=2an+1+a2n-1,证明{bn}为等差数列
数列{an}为等差数列,数列{bn}满足bn=2an+1+a2n-1,证明{bn}为等差数列
数列{an}为等差数列,数列{bn}满足bn=2an+1+a2n-1,证明{bn}为等差数列
如果bn=2a(n+1)+a(2n-1),可解.设an=cn+d(n∈N*)
则bn=2*(c(n+1)+d)+(c(2n-1)+d)=4cn+c+3d
(bn+1)-bn=4c(n∈N*)
{bn}为等差数列
如果bn=2an+a²2n.无法解
应该比较完整!
设an=cn+d
则bn=2*(c(n+1)+d)+(c(2n-1)+d)=4cn+c+d
(bn+1)-bn=4c
{bn}为等差数列