Question

已知数列{an}的前n项和为Sn,且Sn=2an-2(n=1,2,3,L),数列{bn}中,b1=1

已知数列{an}的前n项和为Sn,且Sn=2an-2(n=1,2,3,L),数列{bn}中,b1=1,点P(bn,bn+1)在直线x-y+2=大上.
求1/(b1b2)+1/(b2b3)+…+1/(bnbn+1)的值

Answer 1

s1=a1=2a1-2
a1=2

Sn=2an-2
S(n-1)=2a(n-1)-2
两式相减得
an=2an-2a(n-1)
an=2a(n-1)
an/a(n-1)=2
所以an是以2为公比的等比数列
an=a1q^(n-1)
=2*2^(n-1)
=2^n

x-y+2=0
bn-b(n+1)+2=0
b(n+1)-bn=2
所以bn是以2为公差的等差数列
bn=b1+(n-1)d
=1+2(n-1)
=2n-1

1/[bn*b(n+1)]
=1/[(2n-1)(2n+1)]
=1/2*[1/(2n-1)-1/(2n+1)]
1/(b1b2)+1/(b2b3)+…+1/(bnbn+1)
=1/2*(1-1/3)+1/2*(1/3-1/5)+.........+1/2*[1/(2n-1)-1/(2n+1)]
=1/2*[1-1/3+1/3-1/5+.........+1/(2n-1)-1/(2n+1)]
=1/2*[1-1/(2n+1)]
=1/2*2n/(2n+1)
=n/(2n+1)