计算题1\/2+5\/6+11\/12+19\/20+29\/30+..+9899\/9900=?
原式=1000又1\/100
1\/2+5\/6+11\/12+19\/20+29\/30+...+9701\/9702+9899+9900=???
由此可知上式共有99项 所以 1\/2+5\/6+11\/12+19\/20+29\/30+...+9701\/9702+9899\/9900= 99-(1\/2+1\/6+1\/12+1\/20+1\/30+...+1\/9702+1\/9900)= 99-(1\/2+(1\/2-1\/3)+(1\/3-1\/4)+(1\/4-1\/5)+(1\/5-1\/6)+...+(1\/98-1\/99)+(1\/99-1\/100))= 99-(1\/2+1\/2-...
求教奥数题:1\/2+5\/6+11\/12+19\/20+...+9899\/9900的结果
1\/2+5\/6+11\/12+19\/20+...+9899\/9900 =(1-1\/2)+(1-1\/6)+(1-1\/12)+……+(1-1\/9900)= 99 - (1\/(1*2)+1\/(2*3)+……+1\/(99*100))= 99 - (1-1\/2+1\/2-1\/3+……+1\/99-1\/100)= 99 - (1-1\/100)= 99 - 99\/100 = 98又1\/100 这道题目是通过裂项来...
计算 简便计算 1\/2+5\/6+11\/12+19\/20+……+9890\/9900
1\/2+5\/6+11\/12+19\/20+29\/30+...+9701\/9702+9899\/9900=1\/2+(1-1\/6)+(1-1\/12)+(1-1\/20)+(1-1\/30)+……+(1-1\/9702)+(1-1\/9900) =1\/2+[1-(1\/2-1\/3)]+[1-(1\/3-1\/4)]+[1-(1\/4-1\/5)]+[1-(1\/5-1\/6)]+...
1\/2+5\/6+11\/12+19\/20+29\/30……9899\/990=?
1\/2+5\/6+11\/12+19\/20+29\/30+...+9701\/9702+9899\/9900 =1\/2+(1-1\/6)+(1-1\/12)+(1-1\/20)+(1-1\/30)+……+(1-1\/9702)+(1-1\/9900)=1\/2+[1-(1\/2-1\/3)]+[1-(1\/3-1\/4)]+[1-(1\/4-1\/5)]+[1-(1\/5-1\/6)]+……+[1-(1\/98-1\/99...
1\/2+5\/6+11\/12+19\/20+29\/30+...+9701\/9702+9899\/9900
依此类推,11\/12=1-1\/3+1\/4 19\/20=1-1\/4+1\/5 ...9899\/9900=1-1\/99+1\/100 因此,1\/2+5\/6+11\/12+19\/20+29\/30+...+9701\/9702+9899\/9900 =1\/2+(1-1\/2+1\/3)+(1-1\/3+1\/4)+(1-1\/4+1\/5)...+(1-1\/99+1\/100)=98*1+1\/100 =98又1\/100 或者9801\/100 希...
计算1\/2+5\/6+11\/12+19\/20+29\/30...+9701\/9701+9899\/9900
原式有误,应该为:1\/2+5\/6+11\/12+19\/20+29\/30...+9701\/9702+9899\/9900 =(1-1\/2)+(1-1\/6)+(1-1\/12)+(1-1\/20)+(1-1\/30)+...+(1-1\/9506)+(1-1\/9900)2=1×2,6=2×3,12=3×4,20=4×5,30=5×6,...9701=98×99,9900=99×100,每个分数分解后的较小...
1\/2+5\/6+11\/12+19\/20+29\/30+……+9899\/9900
1\/2+5\/6+11\/12+19\/20+29\/30+……+9899\/9900 =99-(1\/2+1\/6+1\/12+1\/20+1\/30+……+1\/9900)=99-(1-1\/2+1\/2-1\/3+1\/3-1\/4+1\/4-1\/5+1\/5-1\/6+……+1\/99-1\/100)=99-(1-1\/100)=98.01
1\/2+5\/6+11\/12+19\/20+29\/30+...+9701\/9702+9899\/9900
29\/30=1-1\/30 第一项分母2=1*2 第二项分母6=2*3 最后一项分母9900=99*100 所以总共由99项 所以原题=99-(1\/2+1\/6+1\/12+1\/20+1\/30+……+1\/9900 )而 1\/2+1\/6+1\/12+1\/20+1\/30+……+1\/9900 =1-1\/2+1\/2-1\/3+1\/4-1\/5+1\/5-1\/6+……+1\/99-1\/100 =1-1\/...
数学1\\2+5\\6+11\\12+9\\20+...9890\\9900=?
1\\2+5\\6+11\\12+9\\20+...9890\\9900 =100-(1-1\/2+1\/2-1\/3+1\/3-1\/4……+1\/99-1\/00)=100-1+1\/100 =99又99\/100