已知数列{an}满足,a1=1,a2=2,an+2=(an十an+1)/2,n∈N,求{an}的通项公式

Question

如题所述

Answer 1

x^2=(x+1)/2 两根是x1=1,x2=-1/2
所以an通项公式为A×1^n+B×(-1/2)^n A,B为待定系数
a1=A-B/2=1 a2=A+B/4=2 得 A=5/3 B=4/3
an=[5+4×(-1/2)^n]/3

若没有学过特征方程，可如下转换
a[n+2]-a[n+1]=-(a[n+1]-a[n])/2 等比数列
所以a[n+2]-a[n+1]=（-1/2）^n (a[2]-a[1]) =(-1/2)^n n>=0
又a[n+2]+a[n+1]/2=a[n+1]+a[n]/2=................=a[2]+a[1]/2 =5/2 n>=0
两式消去a[n+2]得
(3/2)a[n+1]=5/2-(-1/2)^n n>=0
a[n+1]=5/3-(2/3)(-1/2)^n n>=1
a[n]=(5/3)-(2/3)(-1/2)^(n-1)=(5/3)+(4/3)(-1/2)^n

Answer 2

a(n+2)=[an+a(n+1)]/2
2a(n+2)-a(n+1)-an=0
设2[a(n+2)-xa(n+1)]+y[a(n+1)-xan]=0
则2a(n+2)-(2x-y)a(n+1)-xyan=0
2x-y=1,xy=1
x1=1,y1=1
x2=-1/2,y2=-2

2[a(n+2)-a(n+1)]+[a(n+1)-an]=0,2[a(n+2)+(1/2)xa(n+1)]-2[a(n+1)+(1/2)an]=0
2[a(n+2)-a(n+1)]=-[a(n+1)-an],2[a(n+2)+(1/2)xa(n+1)]=2[a(n+1)+(1/2)an]
相除：
2[a(n+2)-a(n+1)]/[a(n+2)+(1/2)a(n+1)]=-[a(n+1)-an]/[a(n+1)+(1/2)an]
设[a(n+1)-an]/[a(n+1)+(1/2)an]=bn,b1=[a2-a1]/[a2+(1/2)a1]=2/5
2b(n+1)=-bn
bn=b1(-1/2)^(n-1)=(2/5)(-1/2)^(n-1)
[a(n+1)-an]/[a(n+1)+(1/2)an]=bn=(2/5)(-1/2)^(n-1)
a(n+1)-an=a(n+1)(2/5)(-1/2)^(n-1) +an(1/5)(-1/2)^(n-1)
[1-(2/5)(-1/2)^(n-1)]a(n+1)=[1+(1/5)(-1/2)^(n-1)]an
[(-2)^(n-1)-2/5]a(n+1)=[(-2)^(n-1)+1/5]an
a(n+1)/an=[(-2)^(n-1)+1/5]/[(-2)^(n-1)-2/5]
-2a(n+1)/an=[(-2)^(n-1)+1/5]/[(-2)^(n-2)+1/5]

-2an/a(n-1)=[(-2)^(n-2)+1/5]/[(-2)^(n-3)+1/5]
-2a(n-1)/a(n-2)=[(-2)^(n-3)+1/5]/[(-2)^(n-4)+1/5]
-2a(n-2)/a(n-3)=[(-2)^(n-4)+1/5]/[(-2)^(n-5)+1/5]
……
-2a4/a3=[(-2)^2+1/5]/[(-2)^1+1/5]
-2a3/a2=[(-2)^1+1/5]/[(-2)^0+1/5]
-2a2/a1=[(-2)^0+1/5]/[(-2)^(-1)+1/5]
叠乘：
[(-2)^(n-1)]an/a1=[(-2)^(n-2)+1/5]/[(-2)^(-1)+1/5]
=(-2/3)[5(-2)^(n-2)+1]
an=(-2/3)[5(-2)^(n-2)+1]/[(-2)^(n-1)]
=(1/3)[5(-2)^(n-1)-2]/[(-2)^(n-1)]
=(1/3)[5+(-1/2)^(n-2)]
=5/3+(1/3)(-1/2)^(n-2)

Answer 3

已知数列{an}满足a1=1,a2=2,a(n+2)=[an+a(n+1)]/2所以，2a(n+2)-a(n+1)-an=0所以，变形得到2a(n+2)-2a(n+1)=-a(n+1)+an=-[a(n+1)-an]令bn=a(n+1)-an所以，b1=a2-a1=12b(n+1)=-bn所以，b(n+1)=(-1/2)*bn所以，bn=(-1/2)^(n-1)所以，a(n+1)-an=(-1/2)^(n-1)所以，an=∑ [(-1/2)^(n-2)] + a1=-2*(1-(-1/2)^n)/(1-(-1/2)) +1=4((-1/2)^n -1)/3 +1=(4/3)*(-1/2)^n -1/3