设数列{An}满足An+1=An²-NAn+1,n=1,2,3...；当a1=2时,求出a2,a3,a4,a5

来源：学生作业帮编辑：神马作文网作业帮分类：数学作业时间：2024/11/17 04:36:41

设数列{An}满足An+1=An²-NAn+1,n=1,2,3...；当a1=2时,求出a2,a3,a4,a5；并由此猜测出{An}的一个

a2=2^2-1*2+1=3
a3=3^2-2*3+1=4
a4=4^2-3*4+1=5
a5=5^2-4*5+1=6
猜测an=n+1
以下用数学归纳法证明
由a1=2=1+1知n=1时an=n+1成立
设n=k(k属于正整数)时an=n+1成立即ak=k+1
则当n=k+1时,因为a(n+1)=an²-n*an+1,
所以a(k+1)=ak²-k*(k+1)+1
=(k+1)²-k*(k+1)+1
=k²+2k+1-k²-k+1
=k+2
综上,an=n+1成立

设数列｛an｝满足a1+2a2+3a3+.+nan=n(n+1)(n+2) 数列an满足a1+2a2+3a3+...+nan=(n+1)(n+2) 求通项an 已知数列(an)满足a1=1,an+1=2an/an+2(n∈N*) 求a2,a3,a4,a5 猜想数列(an)的通项公已知数列{an}满足a1+a2+a3+...+an=n^2+2n.(1)求a1,a2,a3,a4 若数列{an}满足a1+2a2+3a3+~~+nan=n(n+1)(2n+1),则an= 已知数列(an)满足a1+2a2+3a3+...+nan=n(n+1)(n+2)求an 已知在数列{an}中,a1=1,nan+1=2(a1+a2+a3+...+an)(n∈N*)(1)求a2,a3,a4（2 设数列｛an}的前n项和为Sn,a1=1,Sn=nan-2n(n-1) (1)求a2,a3,a4,并求出数列{an}的通设数列{an}的前n项和为Sn,且a1=1,Sn=nan-2n(n-1) (1)求a2,a3,a4,并求出数列{an}的已知数列{an}满足a1+2a2+3a3+…+nan=n(n+1)(n+2),则a1+a2+a3+…+an=多少? 在数列an中,a1=1/2 an+1=3an/an+3 求a2 a3 a4 a5? 已知数列{an}满足:a1=1,且an-an-1=2n,求(1)a2,a3,a4.(2)求数列{an}的通项an