Question

設(shè)函數(shù)f（x）=x^m+ax的導(dǎo)數(shù)f′（x）=2x+3，則數(shù)列{

1

f(n)+2

}（n∈N^*）的前n項(xiàng)和是（　�。�

• A．

  2n

  n+1

• B．

  n

  2(n+2)

• C．

  n-1

  n+1

• D．

  2(n+1)

  n+2

Answer 1

分析：由f（x）=x^m+ax的導(dǎo)數(shù)f'（x）=mx^m-1+a=2x+3，先求出f（x）=x²+3x，設(shè)a_n=

1

f(n)+2

=

1

n2+3n+2

=

1

(n+1)(n+2)

=

1

n+1

-

1

n+2

，由此能求出數(shù)列{

1

f(n)+2

}（n∈N^*）的前n項(xiàng)和．

解答：解：∵f（x）=x^m+ax的導(dǎo)數(shù)f'（x）=mx^m-1+a=2x+3，
∴m=2，a=3，
∴f（x）=x²+3x，
設(shè)a_n=

1

f(n)+2

，
∴則a_n=

1

f(n)+2

=

1

n2+3n+2

=

1

(n+1)(n+2)

=

1

n+1

-

1

n+2

，
∴數(shù)列{

1

f(n)+2

}（n∈N^*）的前n項(xiàng)和
S_n=a₁+a₂+…+a_n
=(

1

2

-

1

3

)+(

1

3

-

1

4

)+…+(

1

n+1

-

1

n+2

)
=

1

2

-

1

n+2

=

n

2(n+2)

．
故選B．

點(diǎn)評(píng)：本題考查數(shù)列的性質(zhì)和應(yīng)用，是基礎(chǔ)題．解題時(shí)要認(rèn)真審題，注意導(dǎo)數(shù)的性質(zhì)和應(yīng)用．