a1=2p,an=2p-p^2/an-1p为常数求a2a3a4猜想an的通项公式,并证明···························有分追加········吖········是a(n-1)

Question

a1=2p,an=2p-p^2/an-1 p为常数求a2a3a4 猜想an的通项公式,并证明
···························有分追加········吖········是a(n-1)

Answer 1

a1=2p
a2=2p-p²/a1=2p-p²/2p=3/2*p
a3=2p-p²/a2=2p-2p²/3p=4/3*p
a4=2p-p²/a3=2p-3p²/4p=5/4*p
猜想an=（n+1）/n*p（n∈正整数）,且a1,a2,a3,a4均成立
假设an成立,那么a（n+1）=2p-p²/an=2p-p²n/（n+1）p=2p-np/（n+1）=（n+2）/（n+1）*p
所以a（n+1）也成立
根据数学归纳法,所以an=（n+1）/n*p（n∈正整数）