an=1/[n^1/2+(n+1)^1/2],Sn=?

An={n （1 已知数列｛an｝中a1=6,且an-an-1=(an-1/n)+n+1(n属于N*,n≥2）,求an 数列{an}满足a1=1 an+1=2n+1an/an+2n 设an=(1/n+1)+(1/n+2)+(1/n+3)+...+1/2n,则an+1-an等于? 数列｛an},a1=1,an+1=2an-n^2+3n,求｛an｝. an=n(n+1) 求sn 已知an=5n(n+1)(n+2)(n+3),求数列{an}的前n项和Sn 数列{an}中,a1=1,an+1/an=n/n+2,求an 数列{an}中,a1=2,an+1=4an-3n+1,求证{an-n}是等比数列 4an中n为下标an+1中n+1为下标an-n中an的n为下标 已知数列an中,a1=1,an/an-1=n+1/n,n大于等于2,求an 设bn=(n-1)/(an-2),(n大于等于2),an=n^a-n+2,且b(n+1)+b(n+2)+...b(2n+1) a1=1,a(n+1)=2an+n^2+2n +2 求an 数列an,an=(2n-1)+1/【3n(n+1)】,求Sn a1=1/2,Sn=n^2an-n(n-1),求Sn和an 已知数列{An}满足A1=0.5,A1+A2+…+An=n^2An(n∈N*),试用数学归纳法证明:An=1/n(n+1) A1=1,A(n+1)/An=(n+2)/n,求An? an=(3n-2)(1/2)^(n-1),求{an}前n项和 求an的前n项和Sn an=(2n-1)*2^(n-1)