求I y+1 I -I y+2I +I y+3I -I y+4I+。。。I y+2007I-I y+2008 的值 （已知条件如下）

已知3y=-π，计算上述式子

因为y约为－1.几

所以y+1小于0，其余都大于0

去绝对值：-y-1+(-y-2+y+3-y-4+...+y+2007-y-2008)

=-y-1+(-2+3-4+5-6+...-2006+2007-y-2008) (y被约去了）

=-y-1+1*1003-y-2008

=-2y-6

=(2π/3)-6

因为3y=-π 所以 -2<y<-1

I y+1 I -I y+2I +I y+3I -I y+4I+。。。I y+2007I-I y+2008 |

= -( y+1 ) -( y+2) +( y+3) -( y+4)+。。。( y+2007)-( y+2008 )

=-1-2+3-4+........+2007-2008

=-1-1+(1-2)+(3-4)+........+(2007-2008)

=-1-1-.....-1 (1006个 -1）

= -1006

Π=3.14159265····，y=-1.01·····那么▏y+1 ▏=-y-1 ,▏y+2 ▏=y+2,▏y+3 ▏=y+3 ```````

原式=-y-1-y-2+y+3-y-4+y+5-``````````+y+2007-y-2008=-y-1+(-y-2+y+3)+(-y-4+y+5)+······+（-y-2006+y+2007)-y-2008=-y-1+1003-y-2008=2Π/3-1006.

y=-π/3

I y+1 I -I y+2I +I y+3I -I y+4I+。。。I y+2007I-I y+2008 |

=π/3-1-(2-π/3)+(3-π/3)-(4-π/3)+……+(2007-π/3)-(2008-π/3)

=2π/3-3-1-1-……-1

=2π/3-3-1003

=2π/3-1006

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