求数列通项(用特征根法):已知a1=1,a2=2,4a(n+2 )=4a(n+1)-a(n)-1求数列通项公式:一定要用特征根法求(1)已知a1=1,a2=2,4a(n+2 )=4a(n+1)-a(n)-1(2)已知a1=1,a2=2,4a(n+2 )=4a(n+1)-a(n)-n-4一定要用特征根法求,(n+2)
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![求数列通项(用特征根法):已知a1=1,a2=2,4a(n+2 )=4a(n+1)-a(n)-1求数列通项公式:一定要用特征根法求(1)已知a1=1,a2=2,4a(n+2 )=4a(n+1)-a(n)-1(2)已知a1=1,a2=2,4a(n+2 )=4a(n+1)-a(n)-n-4一定要用特征根法求,(n+2)](/uploads/image/z/9320008-40-8.jpg?t=%E6%B1%82%E6%95%B0%E5%88%97%E9%80%9A%E9%A1%B9%EF%BC%88%E7%94%A8%E7%89%B9%E5%BE%81%E6%A0%B9%E6%B3%95%EF%BC%89%EF%BC%9A%E5%B7%B2%E7%9F%A5a1%3D1%2Ca2%3D2%2C4a%28n%2B2+%29%3D4a%28n%2B1%29-a%28n%29-1%E6%B1%82%E6%95%B0%E5%88%97%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F%EF%BC%9A%E4%B8%80%E5%AE%9A%E8%A6%81%E7%94%A8%E7%89%B9%E5%BE%81%E6%A0%B9%E6%B3%95%E6%B1%82%281%29%E5%B7%B2%E7%9F%A5a1%3D1%2Ca2%3D2%2C4a%28n%2B2+%29%3D4a%28n%2B1%29-a%28n%29-1%282%29%E5%B7%B2%E7%9F%A5a1%3D1%2Ca2%3D2%2C4a%28n%2B2+%29%3D4a%28n%2B1%29-a%28n%29-n-4%E4%B8%80%E5%AE%9A%E8%A6%81%E7%94%A8%E7%89%B9%E5%BE%81%E6%A0%B9%E6%B3%95%E6%B1%82%2C%EF%BC%88n%2B2%29)
求数列通项(用特征根法):已知a1=1,a2=2,4a(n+2 )=4a(n+1)-a(n)-1求数列通项公式:一定要用特征根法求(1)已知a1=1,a2=2,4a(n+2 )=4a(n+1)-a(n)-1(2)已知a1=1,a2=2,4a(n+2 )=4a(n+1)-a(n)-n-4一定要用特征根法求,(n+2)
求数列通项(用特征根法):已知a1=1,a2=2,4a(n+2 )=4a(n+1)-a(n)-1
求数列通项公式:一定要用特征根法求
(1)已知a1=1,a2=2,4a(n+2 )=4a(n+1)-a(n)-1
(2)已知a1=1,a2=2,4a(n+2 )=4a(n+1)-a(n)-n-4
一定要用特征根法求,(n+2) (n+1)(n)都是下标
求数列通项(用特征根法):已知a1=1,a2=2,4a(n+2 )=4a(n+1)-a(n)-1求数列通项公式:一定要用特征根法求(1)已知a1=1,a2=2,4a(n+2 )=4a(n+1)-a(n)-1(2)已知a1=1,a2=2,4a(n+2 )=4a(n+1)-a(n)-n-4一定要用特征根法求,(n+2)
1
齐次方程
4a(n+2 )-4a(n+1)+a(n)=0的特征方程为4r^2-4r+1=0
解得r1,2=1/2
所以齐次方程的通解为a1(n)=(c1+c2*n)(1/2)^n
特解很容易看出来a*(n)=-1
所以a(n)=a1(n)+a*(n)=(c1+c2*n)(1/2)^n-1
把a1=1,a2=2带进去解得c1=-4,c2=8
所以a(n)=(8n-4)(1/2)^n-1
2
齐次方程的通解跟1是一样的,a1(n)=(c1+c2*n)(1/2)^n
设特解为a*(n)=an^2+bn+c
带入4a(n+2 )=4a(n+1)-a(n)-n-4解得
a=c=0,b=-1
所以a(n)=a1(n)+a*(n)=(c1+c2*n)(1/2)^n-n
把a1=1,a2=2带进去解得c1=-8,c2=12
所以a(n)=(12n-8)(1/2)^n-n
1就就尽快