Question

已知數(shù)列a_n=2n，前n項(xiàng)和為S_n，若數(shù)列{

1

Sn

}的前n項(xiàng)和為T_n，則T₂₀₁₂的值為（　�。�

• A．

  2012

  2011

• B．

  2010

  2011

• C．

  2013

  2012

• D．

  2012

  2013

Answer 1

分析：由已知數(shù)列a_n=2n，可知數(shù)列{a_n}是一個(gè)等差數(shù)列，可求出其前n項(xiàng)和S_n=n²+n．而

1

Sn

=

1

n(n+1)

=

1

n

-

1

n+1

，故可用裂項(xiàng)求和求出T_n．

解答：解：∵數(shù)列a_n=2n，∴數(shù)列{a_n}是一個(gè)等差數(shù)列，∴前n項(xiàng)和S_n=

n(2+2n)

2

=n²+n．
∴

1

Sn

=

1

n(n+1)

=

1

n

-

1

n+1

，
∴T_n=(1-

1

2

)+(

1

2

-

1

3

)+…+(

1

n

-

1

n+1

)=1-

1

n+1

=

n

n+1

．
∴T₂₀₁₂=

2012

2013

．
故選D．

點(diǎn)評(píng)：本題考查了等差數(shù)列求和及裂項(xiàng)求和問(wèn)題，理解其公式及計(jì)算方法是解題的關(guān)鍵．