一道数列题~A1=1,An+1=2An+(-1)n次方1,若Bn=A2n-1-1/3,求证Bn是等比数列2.求An通项公

1.、Bn+1=A2(n+1) -1 -(1/3)=A2n+1 -(1/3)=(2A2n +1)-1/3=[(2*2A2n-1 -1)+1] -1/3=4A2n-1 -4/3
=4Bn
于是Bn+1/Bn=4,{Bn}是等比数列.
2、由1,B1=A1-1/3=2/3,Bn=B1*4^(n-1)=1/3*[(2^(2n-1)]
于是当n=2k-1时,A2n-1=Bn+1/3=1/3*[2^（2n-1）+1]
,由An+1=2An+(-1)n知An+1-An=An+（-1）^n,从而
A2-A1=A1-1; A3-A2=A2+1; A4-A3=A3-1;A5-A4=A4+1……An-An-1=An-1+(-1)^n
将上式相加,得1,当n=2k时,An=A1+A2+A3+.+An-1;
2,当n=2k-1时 ,An=A1+A2+A3+.+An-1 +1
而当n=2k-1时,A2k-1=Bk+1/3=1/3*[2^（2k-1）+1]
故n=2k时,A2k=2A2k-1+(-1)^(2k-1）=2A2k-1 -1=2/3*[2^（2k-1）+1] -1
所以An={1/3*[2^（2n-1）+1] ,n=2k-1
{2/3*[2^（2k-1）+1] -1 ,n=2k.
请君核查,第三小题容哥再想想,我得去吃饭了,输入太慢了,都搞了快1个小时了,呵呵……