题目内容
(06年江苏卷)(14分)
设数列、、满足:,(n=1,2,3,…),证明为等差数列的充分必要条件是为等差数列且(n=1,2,3,…)
解析:证明:必要性,设是{an}公差为d1的等差数列,则
bn+1bn=(an+1an+3) (anan+2)= (an+1an) (an+3an+2)= d1 d1=0
所以bnbn+1 ( n=1,2,3,…)成立。
又cn+1cn=(an+1an)+2 (an+2an+1)+3 (an+3an+2)= d1+2 d1 +3d1 =6d1(常数) ( n=1,2,3,…)
所以数列{cn}为等差数列。
充分性: 设数列{cn}是公差为d2的等差数列,且bnbn+1 ( n=1,2,3,…)
∵cn=an+2an+1+3an+2 ①
∴cn+2=an+2+2an+3+3an+4 ②
①-②得cncn+2=(anan+2)+2 (an+1an+3)+3 (an+2an+4)=bn+2bn+1+3bn+2
∵cncn+2=( cncn+1)+( cn+1cn+2)= 2 d2
∴bn+2bn+1+3bn+2=2 d2 ③
从而有bn+1+2bn+2+3bn+3=2 d2 ④
④-③得(bn+1bn)+2 (bn+2bn+1)+3 (bn+3bn+2)=0 ⑤
∵bn+1bn≥0, bn+2bn+1≥0 , bn+3bn+2≥0,
∴由⑤得bn+1bn=0 ( n=1,2,3,…),
由此不妨设bn=d3 ( n=1,2,3,…)则anan+2= d3(常数).
由此cn=an+2an+1+3an+2= cn=4an+2an+13d3
从而cn+1=4an+1+2an+25d3 ,
两式相减得cn+1cn=2( an+1an) 2d3
因此(常数) ( n=1,2,3,…)
所以数列{an}公差等差数列。