一道小小的数列题数列{an}中.a1=1.a2=4.a(n+2)=2a(n+1)-an+2.求an.PS:a(n+2)表示第n+2项.a(n+1)表示第n+1项
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![一道小小的数列题数列{an}中.a1=1.a2=4.a(n+2)=2a(n+1)-an+2.求an.PS:a(n+2)表示第n+2项.a(n+1)表示第n+1项](/uploads/image/z/1159747-43-7.jpg?t=%E4%B8%80%E9%81%93%E5%B0%8F%E5%B0%8F%E7%9A%84%E6%95%B0%E5%88%97%E9%A2%98%E6%95%B0%E5%88%97%7Ban%7D%E4%B8%AD.a1%3D1.a2%3D4.a%28n%2B2%29%3D2a%28n%2B1%29-an%2B2.%E6%B1%82an.PS%3Aa%28n%2B2%29%E8%A1%A8%E7%A4%BA%E7%AC%ACn%2B2%E9%A1%B9.a%28n%2B1%29%E8%A1%A8%E7%A4%BA%E7%AC%ACn%2B1%E9%A1%B9)
一道小小的数列题数列{an}中.a1=1.a2=4.a(n+2)=2a(n+1)-an+2.求an.PS:a(n+2)表示第n+2项.a(n+1)表示第n+1项
一道小小的数列题
数列{an}中.a1=1.a2=4.a(n+2)=2a(n+1)-an+2.求an.
PS:a(n+2)表示第n+2项.a(n+1)表示第n+1项
一道小小的数列题数列{an}中.a1=1.a2=4.a(n+2)=2a(n+1)-an+2.求an.PS:a(n+2)表示第n+2项.a(n+1)表示第n+1项
∵[a(n+2)-a(n+1)]-[a(n+1)-an]=2
令bn=a(n+1)-an
∴bn为等差数列 又b1=a2-a1=3
∴bn=3+2(n-1)=2n+1
∴a(n+1)=an+2n+1
an=a(n-1)+2(n-1)+1
······
a2=a1+2(2-1)+1
∴a(n+1)=(1+n)n+n+1=(n+1)²
∴an=n² n≥1
d=a2-a1=3
a(n+2)=2a(n+1)-an+2
a(n+2)-a(n+1)=a(n+1)-an+2=3
an=a(n+1)-1
a(n+2)=a(n+1)+a(n+1)-an+2
a(n+2)-a(n+1)=a(n+1)-an+2
[a(n+2)-a(n+1)]-[a(n+1)-an]=2 即数列{a(n+1)-an}成等差数列
设bn=a(n+1)-an
b1=4-1=3
bn=3+2(n-1)=2n+1
a(n+1)-an=2n+1
an=a1+(a2-a1)+(a3-a2)+.....(an-a(n+1))
=1+3+5+.....2n+1
=(1+2n+1)(n+1)/2
=(n+1)(n+1)