数列{an},an=1/[n*2^(n-1)].前N项和为Sn,求证Sn

An=1/n^2 数列求和An=1/n^2 数列(An)求和 数列{an}满足a1=1 an+1=2n+1an/an+2n 数列｛an},a1=1,an+1=2an-n^2+3n,求｛an｝. 数列{an}中,a1=1,an+1/an=n/n+2,求an 设数列{an}中,若an+1 =an+ an+2 (n∈N*),则称数列{an}为“凸数列” .设数列{an}为“凸数列”求第二问证明设数列{an}中,若an+1 =an+ an+2 (n∈N*),则称数列{an}为“凸数列” .设数列{an}为“凸数列”,若a1 =1, 数列｛an｝中,an=-2n+2*(-1)^n,则数列｛an｝的前n项和sn为 已知数列an中,a1=1,an/an-1=n+1/n,n大于等于2,求an 数列{an}满足a1=2,a(n+1)=2an+n+2,求an 数列an满足a1=1/3,Sn=n(2n-1)an,求an 数列{an}中,a1=2,an+1=an+ln(n+1/n),求通项公式 数列{an},a1=3,an*a(n+1)=(1/2)^n,求an 数列an,a1=1.an+1/an=n+2/n,求通项公式, 已知数列｛an｝中,a(n+1)=an+2^n,a1=3,求an 数列an中,若a（ n+1）=an+(2n-1)求an 数列an集合中,a1=2,an+1=an+(2n+1),求an. 设数列{an}中,a1=2,an+1=an+n+1,则通项an=? 数列{an}中,a1=35,an+1-an=2n-1,求an 已知数列an,a1=1 an+1-an=2的n次幂求an