Question

已知數列{a_n}為

1

2

，

1

3

+

2

3

，

1

4

+

2

4

+

3

4

，

1

5

+

2

5

+

3

5

+

4

5

，…．若b_n=

1

an•an+2

，則{b_n}的前幾項和S_n=（　　）

A． 3n2+5n (n+1)(n+2)	B． 2n2+5n (n+1)(n+3)
C． 3n2+2n (n+2)(n+3)	D． 2n2+3n (n+1)(n+2)

Answer 1

an=

1+2+…+n

n+1

=

n(n+1) 2

n+1

=

n

2

．
∴bn=

1

n 2 • n+2 2

=2(

1

n

-

1

n+2

)．
∴S_n=2[(1-

1

3

)+(

1

2

-

1

4

)+(

1

3

-

1

5

)+…+(

1

n-1

-

1

n+1

)+(

1

n

-

1

n+2

)]
=2(1+

1

2

-

1

n+1

-

1

n+2

)
=

3n2+5n

(n+1)(n+2)

．
故選A．