Question

已知{a_n}是首項(xiàng)為1的等差數(shù)列，S_n是{a_n}的前n項(xiàng)和，且S₅=a₁₃，則數(shù)列{

1

anan+1

}的前5項(xiàng)和為（　�。�

A． 10 11

B． 5 11

C． 4 5

D． 2 5

Answer 1

設(shè)等差數(shù)列{a_n}的公差為d，
∵S₅=a₁₃，
∴5×1+

5×4

2

d=1+12d，
解得d=2．
∴a_n=a₁+（n-1）d=1+2（n-1）=2n-1．
∴

1

anan+1

=

1

(2n-1)(2n+1)

=

1

2

(

1

2n-1

-

1

2n+1

)．
∴數(shù)列{

1

anan+1

}的前n項(xiàng)和T_n=

1

2

[(1-

1

3

)+(

1

3

-

1

5

)++…+(

1

2n-1

-

1

2n+1

)]=

1

2

(1-

1

2n+1

)=

n

2n+1

．
∴T₅=

5

2×5+1

=

5

11

．
故選B．