Question

求式子I1/3-1/2I+I1/4-1/3I+I1/5-1/4I+....+I1/10-1/9I=

I1/5-150/559I+I150/559-1/3I-I-1/3I用合理的方法算

Answer 1

I1/3-1/2I+I1/4-1/3I+I1/5-1/4I+....+I1/10-1/9I
=1/2-1/3+1/3-1/4+1/4-1/5+……+1/9-1/10
=1/2-1/10
=2/5

追问

I1/5-150/559I+I150/559-1/3I-I-1/3I用合理的方法算

追答

I1/5-150/559I+I150/559-1/3I-I-1/3I
=150/559-1/5+1/3-150/559-1/3
=-1/5

追问

若实数a、b满足a/IaI+b/IbI=0，则ab/IabI=
1/6+1/12+1/20+1/30+1/42+1/56=

追答

若实数a、b满足a/IaI+b/IbI=0，
a+b=0
a=-b
则ab/IabI=-1
1/6+1/12+1/20+1/30+1/42+1/56
=1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8
=1/2-1/8
=3/8

若实数a、b满足a/IaI+b/IbI=0，
a+b=0
a=-b
则ab/IabI=-1
1/6+1/12+1/20+1/30+1/42+1/56
=1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8
=1/2-1/8
=3/8

追问

若a=2004/2003，b=2003/2002，c=2002/2001，则a、b、c的关系是用>符号连接

追答

a=2004/2003，b=2003/2002，c=2002/2001，
a<b<c (2003分之2004、2002分之2003、2001分之2002）

Answer 2

1/2-1/3+1/3-1/4+1/4-1/5+....+1/9-1/10=1/2-1/10=2/5