设数列a1,a2,a3...,an,...中的每一项都不为0.证明：{an}为等差数列的充分必要条件是：对任何n属于N,都有1/a1*a2+1/a2*a3+...1/an*an+1=n/a1*an+1

设数列a1,a2,a3...,an,...中的每一项都不为0.证明：{an}为等差数列的充分必要条件是：对任何n属于N,都有1/a1*a2+1/a2*a3+...1/an*an+1=n/a1*an+1
设数列a1,a2,a3...,an,...中的每一项都不为0.证明：{an}为等差数列的充分必要条件是：对任何n属于N,都有1/a1*a2+1/a2*a3+...1/an*an+1=n/a1*an+1

设数列a1,a2,a3...,an,...中的每一项都不为0.证明：{an}为等差数列的充分必要条件是：对任何n属于N,都有1/a1*a2+1/a2*a3+...1/an*an+1=n/a1*an+1
先证必要性
若为等差数列,则a1=a 差为d
1/(a1a2)+1/(a2a3)+……1/(anan+1)=1/a(a+d)+1/(a+d)(a+2d)+……1/(a+(n-1)d)(a+nd)
裂项得=(1/d)*(1/a-1/(a+d)+1/(a+d)……-1/(a+(n-1)d)+1/(a+(n-1)d)-1/(a+nd)=(1/d)*(1/a-1/(a+nd))=n/a(a+nd)=n/a1*an+1
再证充分性由
1/a1*a2+1/a2*a3+...1/an*an+1=n/a1*an+1
1/a1*a2+1/a2*a3+...1/an-1*an=n/a1*an
两式相减得
1/(an*an+1)=n/a1*an+1-n/a1*an
-->nan-(n-1)an+1=a
又可得(n-1)an-1-(n-2)an=a
再两式相减得(2n-2)an-(n-1)(an-1+an+1)=0
2an-an-1-an+1=0
-->an+1-an=an-an-1得证