Question

已知數(shù)列{a_n}為首項a₁≠0，公差為d≠0的等差數(shù)列，求S_n=

1

a1a2

+

1

a2a3

+…+

1

anan+1

．

Answer 1

分析：由等差數(shù)列的性質(zhì)可得，

1

anan+1

=

1

d

(

1

an

-

1

an+1

)，利用裂項求和即可．

解答：解：由等差數(shù)列的性質(zhì)可得，

1

anan+1

=

1

d

(

1

an

-

1

an+1

)
∴Sn=

1

d

[(

1

a1

-

1

a2

)+(

1

a2

-

1

a3

)+…+(

1

an

-

1

an+1

)]=

1

d

(

1

a1

-

1

an+1

)=

1

d

an+1-a1

a1an+1

=

n

a1(a1+nd)

．

點(diǎn)評：本題主要考查數(shù)列求和的裂項法、等差數(shù)列的性質(zhì)及前n項和公式．考查學(xué)生的運(yùn)算能力．