1/4+1/12+1/24+1/40……+1/19800=?
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1/4+1/12+1/24+1/40……+1/19800=?1/4+1/12+1/24+1/40……+1/19800=?1/4+1/12+1/24+1/40……+1/19800=?1/4+1/12+1
1/4+1/12+1/24+1/40……+1/19800=?
1/4+1/12+1/24+1/40……+1/19800=?
1/4+1/12+1/24+1/40……+1/19800=?
1/4+1/12+1/24+1/40+……+1/19800
=(1/2)*[1/2+1/6+1/12+1/20+...+1/9900]
=(1/2)*[1/1*2+1/2*3+1/3*4+1/4*5+...+1/99*100]
=(1/2)*[1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+...+1/99-1/100]
=(1/2)*[1-1/100]
=(1/2)*(99/100)
=99/200
加油!
1/4+1/12+1/24+1/40+……+1/19800
=(1/2)*[1/2+1/6+1/12+1/20+...+1/9900]
=(1/2)*[1/1*2+1/2*3+1/3*4+1/4*5+...+1/99*100]
=(1/2)*[1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+...+1/99-1/100]
=(1/2)*[1-1/100]
=(1/2)*(99/100)
=99/200
1/4+1/12+1/24+1/40……+1/19800=?
1/4+1/12+1/24+1/40+……+1/19800=?
4分之1+1/12+1/24+1/40…+1/19800
1+2+3+4……+20+21+23+24……+40的简便算法
有一列数4,6,1,10,12,9,18,24,25,28,48,49,40,96,91,54…………你能看出规律吗
1-2/3-1/6-1/12-1/24-1/48-……1/768
1.(1/3是三分之一,依此类推)1/3+1/6+1/10+1/15+1/21+1/28+1/36+1/452.3/2-5/6+7/12-9/20+11/30-13/423.(括号不要打,我写只是为了分辨)1+(1/2+1)+(1/1+2+3)+……+（1/1+2+3+……+10）4.1/4+1/12+1/24+1/40+……+1/19800,第二题是假
1/3+1/6+1/12+1/24+……+1/3*2^n-1+……= 同上
1/3+1/6+1/12+1/24+……1/96简便计算
1/3-1/6-1/12-1/24……-1/384 怎么计算?
把从一开始的若干个自然数排成下面的形状,那么第10行左起第1个数是12 3 4 5 6 7 8 910 11 12 13 14 15 1617 18 19 20 21 22 23 24 25………………………………………………………………
1…2…6…24…120…?
数学分数计算题3/4×（2.5-1/4）+5/12÷0.25原式=…………………………………………………………
1/4+1/12+1/24+1/40+1/84+1/112+1/144+1/180
3,1,6,2,12,3,24,4,（ ）,（ ）,…… 按规律
1/2+(2/3+1/3)+(3/4+2/4+1/4)+……+(39/40+38/40+……2/40+1/40)
巧算1/4+1/12+1/24+1/40+1/60+1/84+1/112
1+2+3+4+5+6+7+8+9+10+12+13+14+……………………………………………………+9997+9998+9999+10000=?