设数列{An}{Bn}{Cn}满足:Bn:An-A(n+2),Cn=An+2A(n+1)+3A(n+2)(n=1,2,3.)证明{An}为等差数列的充分必要条件是{Cn}为等差数列且Bn小于等于B(n+1)(n=1,2,3.)

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设数列{An}{Bn}{Cn}满足:Bn:An-A(n+2),Cn=An+2A(n+1)+3A(n+2)(n=1,2,3.)证明{An}为等差数列的充分必要条件是{Cn}为等差数列且Bn小于等于B(n+1)(n=1,2,3.)

设数列{An}{Bn}{Cn}满足:Bn:An-A(n+2),Cn=An+2A(n+1)+3A(n+2)(n=1,2,3.)证明{An}为等差数列的充分必要条件是{Cn}为等差数列且Bn小于等于B(n+1)(n=1,2,3.)
设数列{An}{Bn}{Cn}满足:Bn:An-A(n+2),Cn=An+2A(n+1)+3A(n+2)(n=1,2,3.)
证明{An}为等差数列的充分必要条件是{Cn}为等差数列且Bn小于等于B(n+1)(n=1,2,3.)

设数列{An}{Bn}{Cn}满足:Bn:An-A(n+2),Cn=An+2A(n+1)+3A(n+2)(n=1,2,3.)证明{An}为等差数列的充分必要条件是{Cn}为等差数列且Bn小于等于B(n+1)(n=1,2,3.)
先证明必要性:
{An}为等差数列,即A(n+1)-An=d,那么Bn=-2d,Bn=B(n+1);
C(n+1)-Cn=(An+1-An)+2(An+2-An+1)+3(An+3-An+2)=6d,{Cn}也是等差数列;
得证;
再证明充分性:
Bn小于等于B(n+1),
即B(n+1)-Bn=(An+1-An+3)-(An-An+2)=(An+1-An)-(An+3-An+2)>=0,
即An+1-An>=An+3-An+2(n=1,2,3.);
再由{Cn}(n=1,2,3.)为等差数列,
即C(n+1)-Cn=(An+1-An)+2(An+2-An+1)+3(An+3-An+2)=k为一常数;
又An+1-An>=An+3-An+2,即易得:
6(An+3-An+2)=

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