数列an是首次为1的正数列,且(an+1)²/n - an²/n+1 + (an+1*an)/(n+1)n =0,求通项公式数列an是首次为1的正数列,且(an+1)²/n - an²/(n+1) + (an+1*an)/(n+1)n =0,求通项公式
问题描述：
数列an是首次为1的正数列,且(an+1)²/n - an²/n+1 + (an+1*an)/(n+1)n =0,求通项公式
数列an是首次为1的正数列,且(an+1)²/n - an²/(n+1) + (an+1*an)/(n+1)n =0,求通项公式
答
(an+1)²/n - an²/(n+1) + (an+1*an)/(n+1)n =0
方程两边乘以(n+1)n得
(n+1)(an+1)²+an+1*an-nan²=0
[(n+1)an+1-nan](an+1+an)=0
n+1an+1=nan或an+1=-an
an是首项为1的正数列
an+1=-an舍去
n+1an+1/nan=1
nan/n-1an-1
……
2a2/1a1=1
相乘得nan/1a1=1
nan=1
an=1/n
答
(an+1)²/n - an²/(n+1) + (an+1*an)/(n+1)n =0,(n+1)(an+1)² + (an+1*an)- nan² =0,[(n+1)a(n+1)-nan]*[a(n+1)+an]=0a(n+1)+an>0(n+1)a(n+1)-nan=0则{nan}是等差数列,公差为0,即为常数列首项为1...