数列题目，关于递推关系

(2)
a[2] / a[1] = 1 / 3
a[3] / a[2] = 2 / 4
a[4] / a[3] = 3 / 5
...
a[n - 1] / a[n - 2] = (n - 2) / n
a[n] / a[n - 1] = (n - 1) / (n + 1)

将左边全部乘起来再乘上a[1]得
a[1] * (a[2] / a[1]) * ... * (a[n] / a[n - 1])
=a[1] / a[1] * a[2] / a[2] * ... * a[n - 1] / a[n - 1] * a[n]
=a[n] = 2 * (1 / 3) * (2 / 4) * (3 / 5) ... * (n - 2) / n * (n - 1) / (n + 1)
= 1 * 2 * 2 * 3 / 3 * 4 / 4 * 5 / 5 ... * (n - 1) / (n - 1) / n / (n + 1)
= 2 * 2 / n ／ (n + 1)
= 4 / [(n + 1) * n]

(3) //sqrt 表示根号下
1 / [sqrt(n + 1) + sqrt(n)]
={1 * [sqrt(n + 1) - sqrt(n)]} / {[sqrt(n + 1) + sqrt(n)] * [sqrt(n + 1) - sqrt(n)]}
=[sqrt(n + 1) - sqrt(n)] / [(n + 1) - n]
=[sqrt(n + 1) - sqrt(n)] / 1
=sqrt(n + 1) - sqrt(n)

所以 a[n] - a[n - 1] = sqrt(n + 1) - sqrt(n)
a[2] - a[1] = sqrt3 - sqrt2
a[3] - a[2] = sqrt4 - sqrt3
...
a[n] - a[n - 1] = sqrt(n + 1) - sqrt(n)
左边全部加起来
a[n] - a[n - 1]