数列an中,a1=1/2,an+1=nan/【(n+1)*(nan+1)】,前n项和为Sn

题目

数列an中,a1=1/2,an+1=nan/【(n+1)*(nan+1)】,前n项和为Sn
1.设bn=1/nan,求证数列bn是等差数列
2求Sn的表达式

答案

1.
1/A(n+1)=(n+1)(nAn+1)/(nAn)
1/[(n+1)A(n+1)]=(nAn+1)/(nAn)=1+1/(nAn)
B(n+1)=1+Bn
{Bn}是公差为1的等差数列
2.
B1=1/A1=2
Bn=2+(n-1)=n+1=1/(nAn)
An=1/(n(n+1))=(n+1-n)/(n(n+1))=1/n-1/(n+1)
Sn=A1+A2+……+An
=(1/1-1/2)+(1/2-1/3)+……+1/n-1/(n-1)
=1-1/(n-1)
=n/(n-1)