数列{an}中,an+1+an=2n-44(n属于N+),a1=-23 (1)求an.(2)设Sn为{an}的前n项和,求Sn(n+1是a角标)特别是第二问
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![数列{an}中,an+1+an=2n-44(n属于N+),a1=-23 (1)求an.(2)设Sn为{an}的前n项和,求Sn(n+1是a角标)特别是第二问](/uploads/image/z/3198772-28-2.jpg?t=%E6%95%B0%E5%88%97%7Ban%7D%E4%B8%AD%2Can%2B1%2Ban%3D2n%EF%BC%8D44%28n%E5%B1%9E%E4%BA%8EN%2B%29%2Ca1%3D-23+%EF%BC%881%EF%BC%89%E6%B1%82an.%EF%BC%882%EF%BC%89%E8%AE%BESn%E4%B8%BA%7Ban%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%2C%E6%B1%82Sn%EF%BC%88n%2B1%E6%98%AFa%E8%A7%92%E6%A0%87%29%E7%89%B9%E5%88%AB%E6%98%AF%E7%AC%AC%E4%BA%8C%E9%97%AE)
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数列{an}中,an+1+an=2n-44(n属于N+),a1=-23 (1)求an.(2)设Sn为{an}的前n项和,求Sn(n+1是a角标)特别是第二问
数列{an}中,an+1+an=2n-44(n属于N+),a1=-23 (1)求an.(2)设Sn为{an}的前n项和,求Sn(n+1是a角标)
特别是第二问
数列{an}中,an+1+an=2n-44(n属于N+),a1=-23 (1)求an.(2)设Sn为{an}的前n项和,求Sn(n+1是a角标)特别是第二问
(1)an+1+an=2n-44;则an+an-1=2(n-1)-44;……;a2+a1=2-44;将上述式子进行错位相减得an+1-a1=2[n-(n-1+n-2+……+1)]=(3n-n^2)/2则an+1=(3n-n^2)/2-23即an=(-n^2+5n-50)/2
(2)sn=a1+a2+……+an=1/2[-(1^2+2^2+……+n^2)+5(1+2+……+n)-50n]=1/2[-n(n+1)(2n+1)/6+5n(n+1)/2-50n]=(-n^3+6n^2-143n)/6(补充下,1^2+2^2+……+n^2=n(n+1)(2n+1)/6是公式,能直接用,可以用数学归纳法证明)