a1=5,a(n+1)=√(4+an),用数学归纳法证明an为递减数列.

1.a1=3 且 a（n+1）=an+5ana（n+1）2.a1=1 且 a（n+1）=an+2n+13.a1=1 且 a（n+1）=an+1/4n^2-14.a1=1 an=n+1/n*a（n+1）求各题的an 设a1=5,a(n+1)=√（4+an),求该数列的极限, 1.数列{an}满足a(n+1)=an+2^n,且a1=1,求通项公式2.{an}中,a1=4,a(n=1)=5^n * an,求an a1=5,a(n+1)=√(4+an),用数学归纳法证明an为递减数列. a(n+1)=2an/3an+4,a1=1/4,求an 数列｛an｝,a1=2,4a(n+1)-an=2,求通项an 数列{an}中 a1=8 4a(n+1)=an 求通项公式an 数列an,a1=4,an+1=5^n*an,求an用累乘法 a1=1/5,an+a(n+1)=6/(5^n+1),lim(a1+a2+...an )=? a1=1/5,an+a(n+1)=6/【5^(n+1)】lim(a1+a2+...an )=? 数列{an}中,a1=1,a(n+1)=1/4an+√an+1,a7=? 已知数列{an}满足3a(n+1)=2an-4,且a1=1/5,求an 数列{an}满足:a1=2,an=[4a(n-1)]^5,求{an}的通项 在数列{an},a1=1/5,an+a(n+1)=6/5^(n+1),n∈N+,则lim(a1+a2+...+an)的值为? 数列an,a1=4,Sn+S(n+1)=5/3an+1,an数列an,a1=4,Sn+S(n+1)=5（a（n+1））/3,求an 写出下列数列{an}的前5项 a1=1,a(n+1)=an+3 a1=2,a(n+1)=2an a1=3,a2=6,a(n+2)=a(n+1)-an 1、在数列{an}中,a1=1.a(n+1)=3an+2n+1.求an.2、在数列{an}中,a1=-1,a(n+1)=(3an-4)/[(an)-1].求an. 已知数列{an}满足 n∈N 都有a(n+1)=13an-25／an+3 (1)若a1=5,求an (2)若a1=3,求an (3)a1=6 求an （4）当a1取哪些值时,无穷数列an不存在