设数列{an}.{bn}.{cn}满足:bn＝an-an+2,cn＝an+2an+1+3an+2. 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. 证明:必要性:设{a——青夏教育精英家教网—

设数列{an}.{bn}.{cn}满足:bn＝an-an+2,cn＝an+2an+1+3an+2. 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. 证明:必要性:设{an}是公差为d1的等差数列,则bn+1-bn＝(a n+1-an+3)-(an-an+2)＝(a n+1-an)-(a n+3-a n+2)＝d1-d1＝0, ∴bn≤b n+1成立. 又c n+1-cn＝(a n+1-an)+2(a n+2-a n+1)+3(a n+3-a n+2)＝d1+2d1+3d1＝6d1. ∴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≤b n+1. 证法一:∵cn＝an+2a n+1+3a n+2, ① ∴c n+2＝a n+2+2a n+3+3a n+4. ② ①-②,得cn-c n+2＝(an-a n+2)+2(a n+1-a n+3)+3(a n+2-an+4)＝bn+2b n+1+3b n+2, ∴cn-c n+2＝(cn-cn+1)+(c n+1-cn+2)＝-2d2. ∴bn+2bn+1+3bn+2＝-2d2, ③ 从而有bn+1+2bn+2+3bn+3＝-2d2. ④ ④-③,得(bn+1-bn)+2(bn+2-bn+1)+3(bn+3-bn+2)＝0. ⑤ ∵bn+1-bn≥0,bn+2-bn+1≥0,bn+3-bn+2≥0, ∴由⑤得bn+1-bn＝0. 由此不妨设bn＝d3,则an-a n+2＝d3. 由此cn＝an+2an+1+3aa+2＝4an+2an+1-3d3, 从而cn+1＝4an+1+2an+2-3d3＝4a n+1+2an-5d3. 两式相减,得cn+1-cn＝2(an+1-an)-2d3, 因此an+1-an＝(cn+1-cn)+d3＝d2+d3, ∴数列{cn}是等差数列. 证法二:令An＝a n+1-an,由bn≤b n+1,知an-a n+2≤a n+1-a n+3, 从而a n+1-an≥a n+3-a n+2,即An≥A n+2. 由cn＝an+2a n+1+3a n+2,c n+1＝a n+1+2a n+2+3a n+3,得c n+1-cn＝(a n+1-an)+2(a n+2-a n+1)+3(a n+3-a n+2), 即An+2A n+1+3A n+2＝d2时 ⑥ 由此得A n+2+2A n+3+3An+4＝d2. ⑦ ⑥-⑦,得(An-A n+2)+2(A n+1-A n+3)+3(A n+2-An+4)＝0. ⑧ ∵An-A n+2≥0,A n+1-A n+3≥0,A n+2-A n+4≥0, ∴由⑧,得An-A n+2＝0. 于是由⑥,得4An+2A n+1＝An+2A n+1+3A n+2＝d2, ⑨ 从而2An+4A n+1＝4A n+1+2A n+2＝d2. ⑩ 由⑨和⑩,得4An+2A n+1＝2An+4A n+1, 故A n+1＝An, 即a n+2-a n+1＝a n+1-an, ∴数列{cn}是等差数列. 【查看更多】

题目列表(包括答案和解析)

设数列{a_n}、{b_n}、{c_n}满足：b_n＝a_n－a_n_＋2，c_n＝a_n＋2a_n_＋1＋3a_n_＋2(n＝1，2，3，…)，求证：{a_n}为等差数列的充分必要条件是{c_n}为等差数列且b_n≤b_n_＋1(n＝1，2，3，…)．