设数列{an}、{bn}、{cn}满足:bn=an-an+2,cn=an+2an+1+3an+2(n=1,2,3,…),
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设数列{an}、{bn}、{cn}满足:bn=an-an+2,cn=an+2an+1+3an+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≤bn+1(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≤bn+1(n=1,2,3,…)
证明:(必要性)
设是{an}公差为d1的等差数列,则
bn+1-bn=(an+1-an+3)-(an-an+2)=(an+1-an)-(an+3-an+2)=d1-d1=0
所以bn≤bn+1(n=1,2,3,)成立.
又cn+1-cn=(an+1-an)+2(an+2-an+1)+3(an+3-an+2)=d1+2d1+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bn≤bn+1(n=1,2,3,)
∵cn=an+2an+1+3an+2①
∴cn+2=an+2+2an+3+3an+4②
①-②得cn-cn+2=(an-an+2)+2(an+1-an+3)+3(an+2-an+4)=bn+2bn+1+3bn+2
∵cn-cn+2=(cn-cn+1)+(cn+1-cn+2)=-2d2
∴bn+2bn+1+3bn+2=-2d2③
从而有bn+1+2bn+2+3bn+3=-2d2④
④-③得(bn+1-bn)+2(bn+2-bn+1)+3(bn+3-bn+2)=0⑤
∵bn+1-bn≥0,bn+2-bn+1≥0,bn+3-bn+2≥0,
∴由⑤得bn+1-bn=0(n=1,2,3,),
由此不妨设bn=d3(n=1,2,3,)
则an-an+2=d3(常数).
由此cn=an+2an+1+3an+2=cn=4an+2an+1-3d3
从而cn+1=4an+1+2an+2-5d3,
两式相减得cn+1-cn=2(an+1-an)-2d3
因此an+1−an=
1
2(cc+1−cc)+d3=
1
2d2+d3(常数)(n=1,2,3,)
所以数列{an}公差等差数列.
综上所述::{an}为等差数列的充分必要条件是{cn}为等差数列且bn≤bn+1(n=1,2,3,…)
设是{an}公差为d1的等差数列,则
bn+1-bn=(an+1-an+3)-(an-an+2)=(an+1-an)-(an+3-an+2)=d1-d1=0
所以bn≤bn+1(n=1,2,3,)成立.
又cn+1-cn=(an+1-an)+2(an+2-an+1)+3(an+3-an+2)=d1+2d1+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bn≤bn+1(n=1,2,3,)
∵cn=an+2an+1+3an+2①
∴cn+2=an+2+2an+3+3an+4②
①-②得cn-cn+2=(an-an+2)+2(an+1-an+3)+3(an+2-an+4)=bn+2bn+1+3bn+2
∵cn-cn+2=(cn-cn+1)+(cn+1-cn+2)=-2d2
∴bn+2bn+1+3bn+2=-2d2③
从而有bn+1+2bn+2+3bn+3=-2d2④
④-③得(bn+1-bn)+2(bn+2-bn+1)+3(bn+3-bn+2)=0⑤
∵bn+1-bn≥0,bn+2-bn+1≥0,bn+3-bn+2≥0,
∴由⑤得bn+1-bn=0(n=1,2,3,),
由此不妨设bn=d3(n=1,2,3,)
则an-an+2=d3(常数).
由此cn=an+2an+1+3an+2=cn=4an+2an+1-3d3
从而cn+1=4an+1+2an+2-5d3,
两式相减得cn+1-cn=2(an+1-an)-2d3
因此an+1−an=
1
2(cc+1−cc)+d3=
1
2d2+d3(常数)(n=1,2,3,)
所以数列{an}公差等差数列.
综上所述::{an}为等差数列的充分必要条件是{cn}为等差数列且bn≤bn+1(n=1,2,3,…)
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