已知数列{an}满足a1=1,an+a(n+1)=-2n 求证(1)数列{a2n}与{a(2n-1)}均是以2为公差的等差数列;试用n表示和试M=a1a2-a2a3+…+(-1)^(k+1 )*aka(k+1)+...+a(2n-1)a2n-a2na(2n+1)
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![已知数列{an}满足a1=1,an+a(n+1)=-2n 求证(1)数列{a2n}与{a(2n-1)}均是以2为公差的等差数列;试用n表示和试M=a1a2-a2a3+…+(-1)^(k+1 )*aka(k+1)+...+a(2n-1)a2n-a2na(2n+1)](/uploads/image/z/8582681-65-1.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7Ban%7D%E6%BB%A1%E8%B6%B3a1%3D1%2Can%2Ba%EF%BC%88n%2B1%EF%BC%89%3D-2n+%E6%B1%82%E8%AF%81%EF%BC%881%EF%BC%89%E6%95%B0%E5%88%97%7Ba2n%7D%E4%B8%8E%7Ba%EF%BC%882n-1%EF%BC%89%7D%E5%9D%87%E6%98%AF%E4%BB%A52%E4%B8%BA%E5%85%AC%E5%B7%AE%E7%9A%84%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%EF%BC%9B%E8%AF%95%E7%94%A8n%E8%A1%A8%E7%A4%BA%E5%92%8C%E8%AF%95M%3Da1a2-a2a3%2B%E2%80%A6%2B%EF%BC%88-1%EF%BC%89%5E%28k%2B1+%29%2Aaka%28k%2B1%29%2B...%2Ba%282n-1%29a2n-a2na%EF%BC%882n%2B1%29)
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已知数列{an}满足a1=1,an+a(n+1)=-2n 求证(1)数列{a2n}与{a(2n-1)}均是以2为公差的等差数列;试用n表示和试M=a1a2-a2a3+…+(-1)^(k+1 )*aka(k+1)+...+a(2n-1)a2n-a2na(2n+1)
已知数列{an}满足a1=1,an+a(n+1)=-2n 求证(1)数列{a2n}与{a(2n-1)}均是以2为公差的等差数列;
试用n表示和试M=a1a2-a2a3+…+(-1)^(k+1 )*aka(k+1)+...+a(2n-1)a2n-a2na(2n+1)
已知数列{an}满足a1=1,an+a(n+1)=-2n 求证(1)数列{a2n}与{a(2n-1)}均是以2为公差的等差数列;试用n表示和试M=a1a2-a2a3+…+(-1)^(k+1 )*aka(k+1)+...+a(2n-1)a2n-a2na(2n+1)
an+a(n+1)=-2n 推得 a(n+1)+a(n+2)=-2(n+1) 由1-2式得 an-a(n+2)=2
所以M=a2(an-a(n+2).+a(2n)[a(2n-1)-a(2n+1)]=(a2+a4+a6+.+a(2n))*2
因为 an-a(n+2)=2
所以a(2n)-a(2(n-1))=a(2n)-a(2n-2)=2 同理
(1)数列{a2n}与{a(2n-1)}均是以2为公差的等差数列;
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