Question

已知等差數(shù)列{a_n}的公差d≠0，它的前n項(xiàng)和為S_n，若S₅=35，且a₂，a₇，a₂₂成等比數(shù)列．
（I）求數(shù)列{a_n}的通項(xiàng)公式；
（II）設(shè)數(shù)列{

1

Sn

}的前n項(xiàng)和為T_n，求T_n．

Answer 1

（I）設(shè)數(shù)列的首項(xiàng)為a₁，則
∵S₅=35，且a₂，a₇，a₂₂成等比數(shù)列
∴

5a1+10d=35 (a1+6d)2=(a1+d)(a1+21d)

∵d≠0，∴d=2，a₁=3
∴a_n=3+（n-1）×2=2n+1；
（II）S_n=

n(3+2n+1)

2

=n(n+2)
∴

1

Sn

=

1

n(n+2)

=

1

2

(

1

n

-

1

n+2

)
∴T_n=

1

2

(1-

1

3

+

1

2

-

1

4

+

1

3

-

1

5

+…+

1

n

-

1

n+2

)=

1

2

(1+

1

2

-

1

n+1

-

1

n+2

)=

3

4

-

2n+3

2(n+1)(n+2)