已知数列{An}满足A1=5,A2=5,An+1=An+6An-1(n大于等于2,n属于整整数)若数列{An+1+kAn}是等比数列.(n与n+1,n-1全是下标)
(1)求数列{An}的通项公式
(2)求证:当K为奇数是,1/Ak+1/Ak+1<4/3的k+1次方;(k与k+1是下标)
(3)求证:1/A1+1/A2+...1/An<1/2

解：(1)a(n+1)=an+6a(n-1)
特征根方程为x²=x+6 ===> x²-x-6=0 ===> x1=3 x2=-2.
所以有An=αx1^n+βx2^n=(3^n)α+[(-2)^n]β
代入已知A1,A2：
3α-2β=5
9α+4β=5
解得α=1 β=-1.
即an=3^n-(-2)^n.

(2)当k为奇数时ak=3^k+2^k>3^k
1/ak+1/a(k+1)<1/3^k+1/3^(k+1)
`````````````=(1/3^k)(1+1/3)
`````````````=(4/3)/3^k

(3)将n=1,2代入得：a1=5,a2=5,
　　当n>=3的奇数时：3/2<(3/2)^n,即:3/2*2^n<3^n,
那么：3/2*2^n+2^n<3^n+2^n=3^n-(-2)^n,
5/2*2^n<3^n-(-2)^n,
当n>=4的偶数时：7/2<(3/2)^n,即:7/2*2^n<3^n,
那么：7/2*2^n-2^n<3^n-2^n=3^n-(-2)^n,
5/2*2^n<3^n-(-2)^n,
从上知：当n>=3时：始终有：5/2*2^n<3^n-(-2)^n=an,
所以：1/an<2/5*2^n=1/5*2^(-n+1) ,
1/a1+1/a2+……+1/an <1/5+1/5+1/5[1/4+1/8+...+1/2^(n-1)],
因为： 1/4+1/8+...+1/2^(n-1)＝1/2,
所以：1/a1+1/a2+……+1/an <1/5+1/5+1/5*1/2=1/2,
即:1/a1+1/a2+……+1/an <1/2.