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)2,\x0c设bn=a2n+1-a2n-1

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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)2,\x0c设bn=a2n+1-a2n-1

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)2,\x0c设bn=a2n+1-a2n-1
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)2,\x0c设bn=a2n+1-a2n-1

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)2,\x0c设bn=a2n+1-a2n-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

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