Question

【急】(n+1)an+1 2-nan2+an+1*an=0(n=1,2,3~) 求an

说明：an、an+1是角标

{an}为首项为1的正项数列(n+1)an+1²-nan²+an+1*an(n=1,2,3,~~)，求an
打错了，n、n+1才是角标

Answer 1

(n+1)a(n+1)²-nan²+a(n+1)*an=0
两边同除以an²
(n+1)a(n+1)²/an²+a(n+1)/an-n=0
[(n+1)a(n+1)/an-n]*[a(n+1)/an+1]=0
已知{an}为首项为1的正项数列
所以(n+1)a(n+1)/an-n=0
a(n+1)/an=n/(n+1)
推得an/a(n-1)=(n-1)/n
a(n-1)/a(n-2)=(n-2)/(n-1)
........
a2/a1=1/2
叠乘an/a1=1/n
所以an=1/n

Answer 2

因式分解得：[(n+1)a(n+1) - nan][a(n+1) +an]=0
故a(n+1)/an =n/(n+1)或a(n+1)+an=0（舍去，∵数列各项为正）
∴an=[an/a(n-1)] *[a(n-1)/a(n-2)] ...[a2/a1]*a1=[(n-1)/n]*[(n-2)/(n-1)]...[1/2]*a1=a1/n=1/n