观察下列各式，1/1*2=1-1/2 1/2*3=1/2-1/3，1/3*4=1/3-1/4（1）地n个式子是?(2)利用规律算
1/1*2+ 1/2*3+1/3*4+。。。。。。+1/9*10
（3）化简1/n(n+1)+1/(n+1)(n+2)+1/(n+2)(n+3)+.......1/(n+2008)(n+2009)
(4)观察一组数：1/3,1/15,1/35,......,猜想第n个数是什么?(请用含n的式子表达）把他填入求这组数的前n项和：1/3+1/15+1/35+......+()中的括号里，并把这个合式化简。
2、原式=1-1/2+1/2-1/3+1/3-1/4..........+1/9-1/10=1-1/10=9/10
3、原式=1/n-1/(n+1)+1/(n+1)-1/(n+2)..........+1/(n+2008)-1/(n+2009)=1/n-1/(n+2009)
4、第n个：2的2n次方减一 分之一追问
谢谢，能否写一下过程。
追答以上都有过程啊~！
观察下列各式,1\/1*2=1-1\/2 1\/2*3=1\/2-1\/3,1\/3*4=1\/3-1\/4
原式=1\/1*2+1\/2*3+…+1\/11*12 =1-1\/2+1\/2-1\/3+…+1\/11-1\/12 =1-1\/12 =11\/12
观察下列各式,1\/1*2=1-1\/2 1\/2*3=1\/2-1\/3,1\/3*4=1\/3-1\/4
化简成1\/(x-1)-1\/x+1\/(x-2)-1\/(x-1)+…+1\/(x-10)-1\/(x-9)=5\/12 1\/(x-10)-1\/x=5\/12 10\/(x(x-10))=5\/12 x=12
观察下列各式,1\/1*2=1-1\/2 1\/2*3=1\/2-1\/3,1\/3*4=1\/3-1\/4
因为1\/n(n+2)=[1\/n-1\/(n+2)]\/2 所以 1\/2*4+1\/4*6+1\/6*8+...+1\/98*100 =1\/2*(1\/2-1\/4)+1\/2*(1\/4-1\/6)+1\/2*(1\/6-1\/8)+...+1\/2*(1\/98-1\/100)=1\/2*(1\/2-1\/4+1\/4-1\/6+1\/6-1\/8+...+1\/98-1\/100)=1\/2*(1\/2-1\/100)=49\/200.
观察下列各式:1\/1*2=1-1\/2,1\/2*3=1\/2-1\/3.1\/3*4=1\/3-1\/4,……
①1\/8*9=1\/8-1\/9 ②1\/n(n+1)=1\/n-1\/(n+1) (n是正整数）(2)由以上的几个式子及你所找到的规律计算：1\/1*2+1\/2*3+1\/3*4+…+1*2010*2011 =1-1\/2011 =2010\/2011
观察下列各式:1\/1×2=1—1\/2,1\/2×3—1\/2—1\/3,1\/3×4=1\/3—1\/4...
（1）1\/8×9=1\/8-1\/9 1\/n（n+1）=1\/n-1\/（n+1）（2）1\/1×2+1\/2×3+1\/3×4+...+1\/2007×2008 =1-1\/2+1\/2-1\/3+1\/3-1\/4+……+1\/2007-1\/2008 =1-1\/2008=2007\/2008
...等式:1\/1*2=1-1\/2,1\/2*3=1\/2-1\/3,1\/3*4=1\/3-1\/4,将以上三个等式两边...
观察下列等式：1\/1*2=1-1\/2,1\/2*3=1\/2-1\/3,1\/3*4=1\/3-1\/4,将以上三个等式两边分别相加得：1\/1*2+1\/2*3+1\/3*4=1-1\/2+1\/2-1\/3+1\/3-1\/4=1-1\/4=3\/4.（1）猜想并写出：1\/n(n+1)=1\/n-1\/(n+1);(2)直接写出下列各式的计算结果 1\/1*2+1\/2*3+1\/3*4+...
观察下列各式:1\/1×2=1-1\/2;1\/2×3=1\/2-1\/3;1\/3×4=1\/3-1\/4,···
解：(1) 1\/n（n+1）=1\/n-1\/(n+1)(2)1\/2+1\/6+1\/12+···+1\/24=(1-1\/2)+(1\/2-1\/3)+(1\/3-1\/4)+...+(15-1\/16)=1-1\/2+1\/2-1\/3...+1\/15-1\/16 =1-1\/16 =15\/16
观察下列各式1\/1x2=1-1\/2,1\/2x3=1\/2-1\/3,1\/3x4=1\/3-1\/4……
(1)根据以上的式子填写下列各题 ①1\/9×10=( 1\/9 -1\/10 )②1\/n(n+1)=[ 1\/n -1\/(n+1) ] n是正整数 (2)由以上几个式子能找到的规律计算：1\/1×2+1\/2×3+1\/3×4+……+1\/2011×2012 =1-1\/2+1\/2-1\/3+1\/3-1\/4+……+1\/2011-1\/2012 =1-1\/2012 =2011\/2012 ...
...等式:1\/1*2=1-1\/2,1\/2*3=1\/2-1\/3,1\/3*4=1\/3-1\/4,将以上三个等式两边...
解：这是分式相消的思想！观察结果：注意：1\/1*2 = (2-1)\/1*2 = 2\/1*2 - 1\/1*2 = 1\/1 - 1\/2 1\/2*3 = (3-2)\/2*3 = 3\/2*3 - 2\/2*3 = 1\/2 - 1\/3 1\/3*4 = (4-3)\/3*4 = 4\/3*4 - 3\/3*4 = 1\/3 - 1\/4 上述各式相加：左边=1\/1*2+1\/2*3+...
观察下列等式1\/1×2=1-1\/2,1\/2×3=1\/2-1\/3,1\/3×4=1\/3-1\/4,将以上三...
题中有个地方有错：（2）1\/1×2+1\/2×3+1\/3×4+...+1\/n（n-1）= 如果用省略号，默认数列满足通式，显然你这儿n-1小于n，格式不对，给你改成（2）1\/1×2+1\/2×3+1\/3×4+...+1\/[(n-1)n]计算的 (1) 1\/[n(n+1)]= 1\/n-1\/(n+1)(2) 1\/1×2+1\/2×3+...
