Question

已知数列〔an〕满足a1=0.a2=2.且对任意m,n属于正整数都有A2m-1+A2n-1=2Am+n-1+2(m-n)

Answer 1

因为Bn=A(2n+1)-A(2n-1)

所以B(n+1)-Bn=A(2n+3)+A(2n-1)

令m=n+2

则B(n+1)-Bn=A(2m-1)+A(2n-1)

化简得A(2n+3)+A(2n-1)=2A(2n+1)=8

A(2n+3)-A(2n+1)=A(2n+1)-A(2n-1)+8

B(n+1)=Bn+8

所以B为等差数列