Question

計(jì)算

1

3•1!

+

1

4•2!

+

1

5•3!

+…+

1

(n+2)n!

．

Answer 1

分析：根據(jù)題意得通項(xiàng)公式，分子分母同乘（n+1）!，再把a(bǔ)_n拆成兩項(xiàng)相減，然后再求和．

解答：解：由題意得，數(shù)列的通項(xiàng)公式
an=

1

(n+2)n!

=

(n+1)!

n!(n+2)!

=

1

n!

-

1

(n+2)!

，
∴

1

3•1!

+

1

4•2!

+

1

5•3!

+…+

1

(n+2)n!

=（1-

1

3!

）+（

1

2!

-

1

4!

）+（

1

3!

-

1

5!

）+…+[

1

(n-1)!

-

1

(n+1)!

]+[

1

n!

-

1

(n+2)!

]
=1+

1

2!

-

1

(n+1)!

-

1

(n+2)!

=

3

2

-

n+3

(n+2)!

點(diǎn)評(píng)：本題用了裂項(xiàng)相消法求和，通項(xiàng)中含有階乘時(shí)通常分子分母同乘另一個(gè)階乘后再裂項(xiàng)，求和時(shí)消不去的項(xiàng)是前后對(duì)稱的．