斐波那契数列an满足a1=1,a2=2,求证：任意n为正整数,都有（an+1）^(1/n)≥1+1/((an)^(1/n))

数列{An}满足a1=1/2,a1+a2+..+an=n方an,求an 斐波那契数列an满足a1=1,a2=2,求证：任意n为正整数,都有（an+1）^(1/n)≥1+1/((an)^(1/n)) 已知数列an满足an=1+2+...+n,且1/a1+1/a2+...+1/an 数列an满足a1=1/2,a1+a2+a3……an=n^2an,则an 已知数列an'满足a1=1/2,a1+a2+a3+...+an=n^2an,求通项公式 设数列AN满足A1等于1,3(A1+a2+~+AN)=(n+2)an,求通向公式 已知数列{an}满足a1=1,a2=3,an+2=3an+1-2an求an 已知数列满足a1=1/2,an+1=2an/(an+1),求a1,a2已知数列满足a1=1/2,a(n+1)=2an/(an+1),求a1,a2;证明0 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?通项是 斐波那契数列a1=3 a2=7 a(n+2)=a(n+1)+an 求an 已知数列an满足a1=0 a2=1 an=(An-1+An-2)/2 求liman 已知数列an满足a1=0 a2=1 an=(An-1+An-2)/2 求liman 数列An,满足An=An-1 + An-2,其中,A1=1,A2=1,求通项公式. 若数列{an}满足a1=1,a2=5且an+2=an+1-an,求a2000 数列{an}满足a1=3,a2=6,an+2=an+1-an,求a2013 数列{an}满足a1=3,a2=6,an+2=an+1-an,求a2008 几个数列问题.已知数列{an} a1=1,an+1=an/(1+n^2*an) 求an 已知数列{an} 满足a1=1 a1*a2*a3.*an=n^2 求an 设数列{an}满足lg(1+a1+a2+...+an)=n+1,求通项公式an