第二小题若无头绪以数学归纳法入手
若Sn=np/(√a1+√a(n+1))（p是正常数）对正整数n恒成立,所以对S1也恒成立
1/(√a1+√a2) = 1*p/(√a1+√a2) 所以p等于1,不妨设a2 =a1+d
数学归纳法 证明数列an 公差为d,即 an = a(n-1) +d
1# k=1 时,显然 a2=a1+d
2# 若 k= n-1时 an = a(n-1)+d
3# 当 k =n 时
Sn=S(n-1)+1/(√an+√a(n+1)) = (n-1)/(√a1+√an) + 1/(√an+√a(n+1)) = n/(√a1+√a(n+1))
所以 (n-1)[1/(√a1+√a(n+1))-1/(√a1+√an) ]= 1/(√an+√a(n+1)) - /(√a1+√a(n+1))
即 (n-1)[√an-√a(n+1)]/[(√a1+√a(n+1)*(√a1+√an) ] = (√a1 - √an)/[(√a1+√a(n+1))*(√an+√a(n+1)) ]
(n-1)[√an-√a(n+1)]/(√a1+√an) = (√a1 - √an)/[(√an+√a(n+1)]
所以 (n-1)[√an-√a(n+1)][(√an+√a(n+1)] = (√a1 - √an)(√a1+√an)
(n-1)(a(n+1)-an) = (an -a1) = (n-1)d
所以 an+1 - an =d
所以,命题an = a(n-1) +d（n∈Z+) 成立,an 是等差数列