Question

等差數列{a_n}中，a₁=3，公差d=2，S_n為前n項和，求

1

S1

+

1

S2

+…+

1

Sn

．

Answer 1

∵等差數列{a_n}的首項a₁=3，公差d=2，
∴前n項和Sn=na1+

n(n-1)

2

d=3n+

n(n-1)

2

×2=n2+2n(n∈N*)，
∴

1

Sn

=

1

n2+2n

=

1

n(n+2)

=

1

2

(

1

n

-

1

n+2

)，
∴

1

S1

+

1

S2

+…+

1

Sn

=

1

2

[(1-

1

3

)+(

1

2

-

1

4

)+(

1

3

-

1

5

)+…+(

1

n-1

-

1

n+1

)+(

1

n

-

1

n+2

)]
=

3

4

-

2n+3

2(n+1)(n+2)

．