数列{an},{bn}、{cn}满足:bn=an−an+2,cn=an+2an+1+3an+2.若数列{cn}为等差数列,且bn⩽bn+1(n=1,2,3,⋯),求证:数列{an}是等差数列.
又数列{cn}为等差数列,因此cn−cn+2=cn+1−cn+3,
于是bn+2bn+1+3bn+2=bn+1+2bn+2+3bn+3,
即(bn−bn+1)+2(bn+1−bn+2)+3(bn+2−bn+3)=0,
结合bn⩽bn+1(n=1,2,3,⋯)可得{bn}为常数列.
不妨设bn=−2d,于是数列{an}的奇数项和偶数项分别构成以2d为公差的等差数列,因此只需要证明a2−a1=d即可.
根据题意,有c1=a1+2a2+3a3=4a1+2a2+6d,c2=a2+2a3+3a4=2a1+4a2+10d,c3=a3+2a4+3a5=4a1+2a2+18d,
而2c2=c1+c3,
于是不难得到a2−a1=d,
因此数列{an}是等差数列.