Question

已知数列an满足,a1=1,且对任意n属于正整数

都有根号下a1分之一+根号下a2分之一+......+根号下an分之一=2倍根号下anan+1分之一
（1）求a2,a3的值（2）求数列an的通项公式

Answer 1

2/[a(1)a(2)]^(1/2) = 1/[a(1)]^(1/2) + 1/[a(2)]^(1/2),
2/[a(2)]^(1/2) = 1 + 1/[a(2)]^(1/2),
a(2)=1.

2/[a(n)a(n+1)]^(1/2) = 1/[a(1)]^(1/2) + 1/[a(2)]^(1/2) + ... + 1/[a(n)]^(1/2),
2/[a(n+1)a(n+2)]^(1/2) = 1/[a(1)]^(1/2) + 1/[a(2)]^(1/2) + ... + 1/[a(n)]^(1/2) + 1/[a(n+1)]^(1/2)
= 2/[a(n)a(n+1)]^(1/2) + 1/[a(n+1)]^(1/2),

2/[a(n+2)]^(1/2) = 2/[a(n)]^(1/2) + 1,

1/[a(n+2)]^(1/2) = 1/[a(n)]^(1/2) + 1/2.

1/[a(2n-1+2)]^(1/2) = 1/[a(2n-1)]^(1/2) + 1/2,
{1/[a(2n-1)]^(1/2)}是首项为1/[a(1)]^(1/2) = 1, 公差为1/2的等差数列。
1/[a(2n-1)]^(1/2) = 1 + (n-1)/2 = (n+1)/2,
[a(2n-1)]^(1/2) = 2/(n+1),
a(2n-1) = 4/(n+1)^2.

1/[a(2n+2)]^(1/2) = 1/[a(2n)]^(1/2) + 1/2.
{1/[a(2n)]^(1/2)}是首项为1/[a(2)]^(1/2) = 1, 公差为1/2的等差数列。
1/[a(2n)]^(1/2) = 1 + (n-1)/2 = (n+1)/2,
[a(2n)]^(1/2) = 2/(n+1),
a(2n) = 4/(n+1)^2.

a(3) = 4/(2+1)^2 = 4/9.

a(n)的通项公式为，
a(2n-1) = a(2n) = 4/(n+1)^2.