Question

已知数列bn=2n-1,求(1/b1b2)+(1/b2b3)+(1/b3b4)+…+（1/bnbn+1）

Answer 1

1/[b(n)b(n+1)]=1/[(2n-1)(2n+1)]=(1/2)*[1/(2n-1)-1/(2n+1)],

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

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

=(1/2)[1/1-1/(2n+1)]

=(1/2)[2n/(2n+1)]

=n/(2n+1)