斐波那契数列a1=3 a2=7 a(n+2)=a(n+1)+an 求an 能否求出a2005呢?

已知数列{an}满足an^2=a(n+1)an-1(n>=1),且a1=根号2,则与根号(a2005)最接近的自然数是 设数列{an}满足a1+3 a2+3^2 a3+……+3^n-1 an=n/3,a属于N* 求数列{an}的通项 已知数列an,a1=5,a2=2,an=2an-1+3an-2,n大于等于3,能否求出通项,如果能,是多少? 已知数列An满足A1=1,An=3^(n-1)+A(n-1)(n=>2).(1)求A2,A3;(2)证明An（3^n-1 数列的,求通项的已知数列{an}中,a1=1,a2=2,a(n+2)=2/3a(n+1)+1/3an,求an 数列{{an}中,a1=1,a2=2,3a(n+2)=2a(n+1)+an,求数列{an}的通项公式 已知数列{an}满足a1=1;an=a1+2a2+3a3+...+(n-1)a(n-1); 在数列{an}中,已知a1=1 a2=3 a(n+2)=a(n+1)-an n属于N* 求a2008 已知数列｛an｝满足：a1=1,且an-a（n-1）=2n.求a2,a3,a4.求数列｛an｝通项an 设数列{an}满足a1+3a2+3^2a3+.3^n-1×an=n/3,a∈N+. 已知数列{an}{bn}满足a1=1,a2=3,b(n+1)/bn=2,bn=a(n+1)-an,(n∈正整数),求数列 已知数列{an}中满足a1=1,a(n+1)=2an+1 (n∈N*),证明a1/a2+a2/a3+…+an/a(n+1