Question

数列{an}的前n项和为Sn,若Sn=2an-1(n属于N+)则Tn=1/a1a2+1/a2a3+…+1/anan+1的

Answer 1

Sn = 2an -1

n=1

a1=1

an = Sn - S(n-1)

= 2an - 2a(n-1)

an = 2a(n-1)

an/a(n-1) = 2

an/a1 = 2^(n-1)

an = 2^(n-1)

1/[an.a(n+1)] = 1/2^(2n-1)

Tn=1/(a1a2)+1/(a2a3)+…+1/(ana(n+1))

= 1/2^1+ 1/2^3+...+1/2^(2n-1)

= (1/2)[ 1- (1/2)^(2n)]/(1- 1/4)

= (2/3)[1-(1/2)^(2n)]