Question

正项数列an满足[(2n-1)an+1]^2-(2n+1)an^2＝8(n^2)-2 a1＝1

Answer 1

是(2n-1)a(n+1)²吧.步骤较多,最好在电脑上看回答结果.

(2n-1)a(n+1)²-(2n+1)an²=8n²-2=2(2n+1)(2n-1)

等式两边同除以(2n+1)(2n-1)

a(n+1)²/(2n+1)-an²/(2n-1)=2

a(n+1)²/[2(n+1)-1]-an²/(2n-1)=2

a1²/(2×1-1)=1/(2-1)=1

数列{an²/(2n-1)}是以1为首项,2为公差的等差数列.

an²/(2n-1)=1+2(n-1)=2n-1

an²=(2n-1)²

数列是正项数列,an>0

an=2n-1

数列{an}的通项公式为an=2n-1.