Question

a1=2,an=an-1/a(n-1)+1,令sn=a1a2+a2a3+.+anan+1.求sn

Answer 1

a1=2,an=an-1/a(n-1)+1

1/a(n)=1+1/a(n-1),1/a(1)=1/2,所以{1/a(n)}为首项1/a(1)=1/2,公差d=1的等差数列

所以1/a(n)=1/a(1)+(n-1)d=1/2+n-1=n-1/2=(2n-1)/2

所以a(n)=2/(2n-1), a(n)a(n+1)=[2/(2n-1)][2/(2n+1)]=2/(2n-1)-2/(2n+1)

所以

sn=a(1)a(2)+a(2)a(3)+.+a(n)a(n+1)

=(2/1-2/3)+(2/3-2/5)+.+[2/(2n-1)-2/(2n+1)]

=2/1-2/(2n+1)=4n/(2n+1)