数列 an种,a1·a2·a3.·an=n²,求通项公式

数列 an种,a1·a2·a3.·an=n²,求通项公式 数列{an}中,a1+a2+a3···+an=2n+1(n∈N※),求an 设数列【an】满足a1=1,3（a1+a2+a3+······+an）=（n+2）an,求通项an 数列{an}中,a1=1/2,a的n+1项=3an/(an+3),a1·a2+a2·a3+a3·a4+...an·a的n+1项=? 若数列{an}满足a1+3a2+3²a3+···+3^n-1an=(n+1)/3,则an=? 已知数列{an}满足a1+2·a2+2^2·a3+……+2^(n-1)·an=n/2,则an= 若数列{an}满足a1+3a2+3²a3+···+3^n-1an=n/2 则an= 设数列{an}满足a1+2a2+3a3+······nan=n^2求数列｛an}的通项公式 一道数列题,已知a1=1,an+1=2an+2·3^n,写出a1,a2,a3,a4 设数列an满足a1=a2=1,a3=2,且对正整数n都有an·an+1·an+2·an+3=an+an+1+an+2+an+3,,求a1+a2+a3+···+a100的值 以知数列an=1/3n-2,记Sn=a1*a2+a2*a3+a3*a4+·······+an*an+1.求Sn. 在数列{an}中,对任意自然数n∈N*恒有a1+a2+···+an=2n-1,则a1+a2^2+a3^3+···+an^n= 一道数列题~数列{An}中,A1=1,对所有N≥2都有A1·A2·A3……·An=n的平方,则A3+A5=? 高一数学 数列 详细解释一下数列{an}中a1=1,n≥2时,a1·a2·a3……·an=n²,求a3+a5 已知数列{an}满足a1=2,an+1【n+1是下标】=1+an/1-an（n属于N+）,则a1·a2·a3...·a2003= 已知数列{an}满足a1=2,an+1【n+1是下标】=1+an/1-an（n属于N+）,则a1·a2·a3...·a2003=? 已知数列{an}满足a1*a2*a3···a(n-1)*an=n^2,则a10-a4= 已知{an}为等差数列,且a1=2,a1+a2+a3=12.(1)求数列{an}的通项公式（2）令bn=an·3^n,求数列{bn}的前n项和