Question

高二数列*3

1.已知数列{An},A1=1,A2=5,A(n+2)=4A(n+1)-4An,求An

2.已知数列{An},A1=0.5,(An-A(n+1))(n+3)(n+1)=AnA(n+1),求An

3.已知数列{An}{Bn},A1=0,B1=1,A(n+1)=4Bn+2An,2B(n+1)=An+2Bn+2,求An、Bn

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需要清晰的思路和答案呢！

Answer 1

1.原式可花成A(n+2)-2A(n+1)=2[A(n+1)-2An]
令Bn=A(n+1)-2An,则B（n+1）=2Bn,B1=A2-2A1=3
{Bn}是以3为首项，2为公比的等比数列
所以Bn=3*2^n,即A(n+1)-2An=3*2^n
所以A(n+1)-3*(n+1)*2^(n+1)=2[An-3*n*2^n]
令Cn=An-3*n*2^n,则C（n+1）＝2Cn,C1=-3
{Cn}是以-3为首项，2为公比的等比数列
Cn=-3*2^n,即An-3*n*2^n=-3*2^n
所以An＝（3n-1）*2^n
2.两边同除以(n+3)(n+1)AnA(n+1)得
1/A(n+1)-1/An=1/[2(n+1)]-1/[2(n+3)]
1/A(n)-1/A(n-1)=1/[2n]-1/[2(n+2)]
……
1/A2-1/A1=1/4-1/8
将上面n个式子相加得
1/A(n+1)-1/A1=-[1/(n+3)-1/(n+2)+1/3+1/2]/2
所以1/A(n+1）=-[1/(n+3)-1/(n+2)+1/3+1/2]/2+1/A1
=-[1/(n+3)-1/(n+2)+1/3+1/2]/2+2
=29/12-0.5/(n+2)-0.5/(n+3)
所以A(n+1)=1/[29/12-0.5/(n+2)-0.5/(n+3)]
An=1/[29/12-0.5/(n+1)-0.5/(n+2)]
3.这个就佷简单了吧，把其中一个代掉就行了