正项数列{an},a1=1,2 an²=an+1²+an-1²,n≥2,求a6=?
正项数列{an},a1=1,2 an²=an+1²+an-1²,n≥2,求a6=?
题:已知正项数列{An}中,A1=1,2An²=An+1²+An-1²,n≥2,求A6
已知数列{an},满足a1=1/2,Sn=n²×an,求an
数列{an},a1=1,an+1=2an-n^2+3n,求{an}.
已知数列{an}满足 a1=3,an+1=an+3n²+3n+2-1\n(n+1),求an的通项公式
数列{an}中,a1=35,an+1-an=2n-1,求an
数列{an}满足a1=1,且an=an-1+3n-2,求an
已知数列{an}满足an+1=2an+3.5^n,a1=6.求an
正项数列{an}满足an²-(2n-1)an-2n=0,求数列{an}的通项公式
数列A1=1/3,An+1=An+An²/n²求证An>1/2+1/4n
已知数列{an}中,a1=2,an+1=an²+2an(n∈N+)
数列an满足a1=1,当n≥2时an²-(n+2)*an-1*an+2*n*an-1²=0 求通项公