Question

定義數(shù)列{x_n}：x1=1，x_n+1=3

x

3 n

+2

x

2 n

+x_n；數(shù)列{y_n}：yn=

1

1+2xn+3xn2

；數(shù)列{z_n}：zn=

2+3xn

1+2xn+3xn2

；若{y_n}的前n項(xiàng)的積為P，{z_n}的前n項(xiàng)的和為Q，那么P+Q=（　�。�

Answer 1

分析：由x_n+1=3

x

3 n

+2

x

2 n

+x_n，變形為

xn

xn+1

=

1

1+2xn+3 x 2 n

，利用“累乘求積”可得P=

x1

xn+1

．由zn=

2+3xn

1+2xn+3 x 2 n

=

1+2xn+3 x 2 n -1

xn+2 x 2 n +3 x 3 n

=

1

xn

-

1

xn+1

，利用“累加求和”可得Q，進(jìn)而得到P+Q．

解答：解：∵x_n+1=3

x

3 n

+2

x

2 n

+x_n，∴

xn

xn+1

=

1

1+2xn+3 x 2 n

，∴P=y₁y₂•…•y_n=

x1

x2

•

x2

x3

•…•

xn

xn+1

=

x1

xn+1

．
∵zn=

2+3xn

1+2xn+3 x 2 n

=

1+2xn+3 x 2 n -1

xn+2 x 2 n +3 x 3 n

=

1

xn

-

1

xn+1

，∴Q=(

1

x1

-

1

x2

)+(

1

x2

-

1

x3

)+…+(

1

xn

-

1

xn+1

)=

1

x1

-

1

xn+1

．
∵x₁=1，
∴P+Q=

1

xn+1

+1-

1

xn+1

=1．
故選A．

點(diǎn)評(píng)：本題考查了經(jīng)過變形利用“累乘求積”求數(shù)列的乘積、利用“累加求和”求數(shù)列的和的基本技能方法，屬于難題．