已知等差数列{an},满足d>0,an*a(n+1)=4n^2-1,求等差数列an的通项公式

题目

已知等差数列{an},满足d>0,an*a(n+1)=4n^2-1,求等差数列an的通项公式

答案

设A1=a 公差=d
An=a+(n-1)d=a-d+nd A(n+1)=a+nd
AnA(n+1)=(a-d+nd)(a+nd)
=(nd)^2+(2a-d)nd+a^2+a(a-d)
=4n^2-1
d^2=4 (2a-d)d=0 a(a-d)=-1
d=2 a=1
An=2n-1