已知数列{an}是等差数列,且bn=an+a(n-1),求证bn也是等差数列

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已知数列{an}是等差数列,且bn=an+a(n-1),求证bn也是等差数列

已知数列{an}是等差数列,且bn=an+a(n-1),求证bn也是等差数列
已知数列{an}是等差数列,且bn=an+a(n-1),求证bn也是等差数列

已知数列{an}是等差数列,且bn=an+a(n-1),求证bn也是等差数列
设an=a1+(n-1)d,
bn=an+a(n-1)
=a1+(n-1)d+a1+nd
=2a1+(2n-1)d
bn为首项为2a1-d,公差为2d的等差数列

设数列{an}的公差为d
又bn=an+a(n-1),
所以 b(n+1)=a(n+1)+an,
两式相减,
得b(n+1)-bn=a(n+1)-a(n-1)
又a(n+1)= an + d
a(n-1)= an - d
所以b(n+1)-bn=2d
d为常数,
所以bn也是等差数列。

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