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)
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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