已知数列{an}满足a1=1,a2=-13,a(n+2)-2(an+1)+an=2n-6 1)设bn=A(n+1)-An
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已知数列{an}满足a1=1,a2=-13,a(n+2)-2(an+1)+an=2n-6 1)设bn=A(n+1)-An,求数列{bn}的通项公式
2,求n为何值时,an最小
2,求n为何值时,an最小
![已知数列{an}满足a1=1,a2=-13,a(n+2)-2(an+1)+an=2n-6 1)设bn=A(n+1)-An](/uploads/image/z/3813384-48-4.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7Ban%7D%E6%BB%A1%E8%B6%B3a1%3D1%2Ca2%3D-13%2Ca%28n%2B2%29-2%28an%2B1%29%2Ban%3D2n-6+1%29%E8%AE%BEbn%3DA%28n%2B1%29-An)
(1)∵a(n+2)-2a(n+1)+an=(a(n+2)-a(n+1))-(a(n+1)-an)=b(n+1)-bn
∴b(n+1)-bn=2n-6
∴bn-b(n-1)=2(n-1)-6
.
b2-b1=2*1-6
∴上述等式叠加可得:bn-b1=2*(1+2+...+n-1)-6(n-1)
而b1=a2-a1=-13-1=-14
∴bn=b1+2*(1+2+...+n-1)-6(n-1)=-14+2*(1+2+...+n-1)-6(n-1)=n^2-7n-8
(2)由(1)得:a(n+1)-an=n^2-7n-8=(n+1)(n-8)
∴当nan
∴a1>a2>...>a8=a9
∴b(n+1)-bn=2n-6
∴bn-b(n-1)=2(n-1)-6
.
b2-b1=2*1-6
∴上述等式叠加可得:bn-b1=2*(1+2+...+n-1)-6(n-1)
而b1=a2-a1=-13-1=-14
∴bn=b1+2*(1+2+...+n-1)-6(n-1)=-14+2*(1+2+...+n-1)-6(n-1)=n^2-7n-8
(2)由(1)得:a(n+1)-an=n^2-7n-8=(n+1)(n-8)
∴当nan
∴a1>a2>...>a8=a9
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