初一（新运算）数学题，在线等

来源：百度知道 编辑：UC知道 时间：2024/09/22 19:25:23
题目如下：对于任意有理数a，b。现定义新运算@：a@b=（a+1）（b+2）分之1
试求：1@2+3@4+5@6+······+2003@2004的值

看不懂的可以问我~

a@b=（a+1）（b+2）分之1
a@a+1=1/(a+1)(a+3)
=1/2[1/(a+1)-1/(a+3)]
1@2+3@4+5@6+······+2003@2004
=1/2*4+1/4*6＋。。。1/2004*2006
=1/2[1/2-1/4+1/4-...1/2004-1/2006]
=1/2(1/2-1/2006)
=501/2006

1@2+3@4+5@6+······+2003@2004
=1/(2*4)+1/(4*6)+1/(6*8)+1/(8*10)+.....
=1/2(1/2-1/4+1/4-1/6+1/6-1/7.....-1/2006)
=1/2(1002/2006)
=556/2006
=278/1003

1@2+3@4+5@6+······+2003@2004=1/2*4+1/4*6+1/6*8+......+1/2004*2006=1/2(1/2-1/4+1/4-1/6+1/6-1/8+.....+1/2004-1/2006)=1/2(1/2-1/2006)=501/2006
就是裂项相消

1@2+3@4+5@6+······+2003@2004
=1/(2*4) + 1/(4*6) +.....+ 1/(2004*2006)
而 1/n*(n+2)= 1/n - 1/(n+2)
所以,
原式=(1/2 - 1/4) + (1/4 -1/6) +.....+ (1/2004 - 1/2006)
=1/2 - 1/2006
=1003/2006 - 1/2006
=501/1003