Question

在数列{an}中,a1=1/3,并且对任意n属于N*,n≥2都有an×an-1=an-1-an成立

Answer 1

解析：

1、当n≥2时

an×a(n-1)=a(n-1)-an

1/an-1/a(n-1)=1

1/an=1/a(n-1)+1

∴数列{1/an}是以1/a1=3为首项,d=1为公差的等差数列

1/an=3+(n-1)=n+2

an=1/(n+2)

bn=1/an=n+2

2、令数列{an/n}为：Cn

则：Cn=1/n(n+2)=1/2[1/n-1/(n+2)]

C1=1/2(1-1/3)

C2=1/2(1/2-1/4)

C3=1/2(1/3-1/5)

C4=1/2(1/4-1/6)

.

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

Tn=1/2[1/-1/3+1/2-1/4+1/3-1/5+1/4-1/6+1/5-1/7+1/6-1/8+1/7-1/9+.+1/n-1/(n+2)]

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

=3/4-(2n+3)/2(n+1)(n+2)