已知数列{an}满足条件:(n-1)an+1=(n+1)(an-1)(n∈N*),求an通项公式.(前面的an+1中n+1为底数,后面an-1中n为底数)求过程!~~~~~~求详解条件里还有a2=6 忘记打了
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![已知数列{an}满足条件:(n-1)an+1=(n+1)(an-1)(n∈N*),求an通项公式.(前面的an+1中n+1为底数,后面an-1中n为底数)求过程!~~~~~~求详解条件里还有a2=6 忘记打了](/uploads/image/z/2734861-13-1.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7Ban%7D%E6%BB%A1%E8%B6%B3%E6%9D%A1%E4%BB%B6%3A%28n-1%29an%2B1%3D%28n%2B1%29%28an-1%29%28n%E2%88%88N%2A%29%2C%E6%B1%82an%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F.%EF%BC%88%E5%89%8D%E9%9D%A2%E7%9A%84an%2B1%E4%B8%ADn%2B1%E4%B8%BA%E5%BA%95%E6%95%B0%2C%E5%90%8E%E9%9D%A2an-1%E4%B8%ADn%E4%B8%BA%E5%BA%95%E6%95%B0%EF%BC%89%E6%B1%82%E8%BF%87%E7%A8%8B%21%7E%7E%7E%7E%7E%7E%E6%B1%82%E8%AF%A6%E8%A7%A3%E6%9D%A1%E4%BB%B6%E9%87%8C%E8%BF%98%E6%9C%89a2%3D6+++%E5%BF%98%E8%AE%B0%E6%89%93%E4%BA%86)
已知数列{an}满足条件:(n-1)an+1=(n+1)(an-1)(n∈N*),求an通项公式.(前面的an+1中n+1为底数,后面an-1中n为底数)求过程!~~~~~~求详解条件里还有a2=6 忘记打了
已知数列{an}满足条件:(n-1)an+1=(n+1)(an-1)(n∈N*),求an通项公式.(前面的an+1中n+1为底数,后面an-1中n为底数)
求过程!~~~~~~求详解
条件里还有a2=6 忘记打了
已知数列{an}满足条件:(n-1)an+1=(n+1)(an-1)(n∈N*),求an通项公式.(前面的an+1中n+1为底数,后面an-1中n为底数)求过程!~~~~~~求详解条件里还有a2=6 忘记打了
(n-1)an+1=(n+1)(an-1)可化为an+1/(n+1)=an/(n-1)-1/(n-1),(前面的an+1中n+1为底数)
两边同时除以n,可得:an+1/n(n+1)=an/n(n-1)-1/n(n-1),设数列an/n(n-1)为bn,
则bn+1=bn-[1/n(n-1)](n>1),将1/n(n-1)裂项为1/(n-1)-1/n
则bn+1=bn+[1/n-1/(n-1)]………(1)
bn=bn-1+[1/(n-1)-1/(n-2)]…………(2)
……
b3=b2+[1/2-1/1]…………(n-1)
将以上(n-1)个式子相加,得bn+1=b2+(1/n)-1
又b2=a2/2,所以b2=3,因此bn+1=3+(1/n)-1,即bn+1=2+1/n,即bn=1+1/(n-1)(n>1)
所以an=n(n-1)bn=n(n-1)+n,(n>1)
将(n-1)an+1=(n+1)(an-1)中n代1,则可得a1=1
因为a1符合an的通项,所以可写在一起,即an=n(n-1)bn=n(n-1)+n
(n-1)a(n+1)=(n+1)[a(n)-1],n=1,2,...
n=1,0=2[a(1)-1], a(1)=1.
n=2,a(3)=3[a(2)-1].
n=3,2a(4)=4[a(3)-1],a(4)=2[a(3)-1]=2[3a(2)-4],
由于a(2)可取任何值,因此{a(n)}的通项公式不存在哈~~~