题目

bn=3(2-a1)(1-a2)……(1-an) 求bn表达式
打错了，是bn=2(1-a1)(1-a2)……(1-an)
an=n+1的平方分之1

提问时间：2021-03-19

答案

an=1/(n+1)^2
bn=2(1-a1)(1-a2)……(1-an)
b(n-1)=2(1-a1)(1-a2)……[1-a(n-1)]
bn=(1-an)b(n-1)
bn/b(n-1)=1-an=1-1/(n+1)^2=[(n+2)/(n+1)][n/(n+1)]
[(n+1)/(n+2)]bn=[n/(n+1)]b(n-1)
[(n+1)/(n+2)]bn=[n/(n+1)]b(n-1)
=[(n-1)/n]b(n-2)
=[(n-2)/(n-1)]b(n-3)
……
=[3/4]b2
=[2/3]b1
[(n+1)/(n+2)]bn=[2/3]b1
bn=[2/3][(n+2)/(n+1)]b1
=[2/3][(n+2)/(n+1)][2(1-an)]
=(4/3)[(n+2)/(n+1)](1-an)
=(4/3)[(n+2)/(n+1)][1-1/(n+1)^2]
=(4/3)[(n+2)/(n+1)][(n+2)n/(n+1)^2]
=(4/3)n(n+2)^2/(n+1)^3