Question

（2012•茂名二模）在數(shù)列{a_n}中，a_n=

n(n+1)

2

．則
（1）數(shù)列{a_n}的前n項和S_n=

n(n+1)(n+2)

6

；
（2）數(shù)列{S_n}的前n項和T_n=

n(n+1)(n+2)(n+3)

24

．

Answer 1

分析：（1）法一：由an=

n(n+1)

2

=

n2

2

+

n

2

，分組求和可求S_n
（2）由T_n=

C

3 3

+

C

3 4

+C

3 5

+…+

C

3 n+1

+

C

3 n+2

，利用組合式的性質(zhì)可求
法2：（1）：由an=

n(n+1)

2

=

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

6

=

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

6

，代入相消法可求和
（2）由（1）中的Sn=

n(n+1)(n+2)

6

=

[n(n+1)(n+2)(n+3)-(n-1)n(n+1)(n+2)]

24

，可求T_n

解答：解：（1）法一：∵an=

n(n+1)

2

=

n2

2

+

n

2

=

C

2 n+1

        
∴Sn=(

12

2

+

1

2

)+（

22

2

+

2

2

）+…+（

n2

2

+

n

2

）
=

1

2

(12+22+…+n2)+

1

2

(1+2+…+n)
=

1

2

×

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

6

+

1

2

×

n(n+1)

2

=

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

12

+

n(n+1)

4

=

(n+1)(2n2+4n)

12

=

n(n+1)(n+2)

6

=

C

3 n+2

 
（2）T_n=

C

3 3

+

C

3 4

 

+C

3 5

+…+

C

3 n+1

+

C

3 n+2

=

C

4 4

+

C

3 4

+C

3 5

+…+

C

3 n+1

+

C

3 n+2

=C

4 5

+

C

3 5

+

C

3 6

+…+

C

3 n+1

+

C

3 n+2

=

C

4 6

+

C

3 6

+…+

C

3 n+2

=…=

C

4 n+1

+

C

3 n+1

+

C

3 n+2

=

C

3 n+2

+

C

4 n+2

=

C

4 n+3

=

n(n+1)(n+2)(n+3)

24

=

C

4 n+3

法2：（1）：∵an=

n(n+1)

2

=

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

6

=

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

6

∴S_n=

1

6

[（1×2×3-0×1×2）+（2×3×4-1×2×3）+（3×4×5-2×3×4）+…+n（n+1）（n+2）-（n-1）n（n+1）]

=

n(n+1)(n+2)

6

=

C

3 n+2

（2）∵Sn=

n(n+1)(n+2)

6

=

n(n+1)(n+2)[(n+3)-(n-1)]

24

=

[n(n+1)(n+2)(n+3)-(n-1)n(n+1)(n+2)]

24

T_n=

1

24

[1×2×3×4-0×1×2×3）+（2×3×4×5-1×2×3×4）+…+n（n+1）（n+2）（n+3）-n（n-1）（n+1）（n+2）]=

n(n+1)(n+2)(n+3)

24

故答案為：

n(n+1)(n+2)

6

，

n(n+1)(n+2)(n+3)

24

點評：本題主要考查了數(shù)列和的求解，解題的關(guān)鍵是熟練應用組合式的性質(zhì)并能靈活變形．