{an}中,a1=1/2,an+1=nan/(n+1)(nan+1),n∈正整数,设bn=1/nan,求证(1){bn}是等差数列(2)Sn的表达式

题目

{an}中,a1=1/2,an+1=nan/(n+1)(nan+1),n∈正整数,设bn=1/nan,求证(1){bn}是等差数列(2)Sn的表达式

答案

(1)a(n+1)=nan/(n+1)(nan+1),
移项,(n+1)a(n+1)=nan/(nan+1)
两边取倒数,1/(n+1)a(n+1)=1+1/nan
bn=1/nan,所以b(n+1)-bn=1,b1=1/(1/2)=2
即bn=1+n,为等差数列
(2)
an=1/n(n+1)=1/n -1/(n+1)
Sn=1-1/2+1/2-1/3……-1/(n+1)
=1-1/(n+1)