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+1/1
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
原式=1/4*[1/1+1/(1+2)+1/(1+2+3)……+1/(1+2+3+……+99)]
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
1/4+1/12+1/24+1/40+1/84+1/112+1/144+1/180
巧算1/4+1/12+1/24+1/40+1/60+1/84+1/112
简算!4分之1+12分之1+24分之1+40分之1+60分之1
4/1+12/1+24/1+40/1+.19800/1
如何巧算1/4+1/12+1/24+1/40+.+1/19800
巧算1/4+1/12+1/24+1/40+1/60+1/84+1/112求详细过程
1/4+1/12+1/24+1/40+1/60+1/84+1/112能不能说清楚些
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
1/12+1/24+1/40+1/60+1/84+1/112+1/144+1/180
计算:4分之1+12分之1+24分之1+40分之1+60分之1+...+180分之1=?
4分之1+12分之1+24分之1+40分之1+60分之1+...+180分之1=?快
1/4+1/12+1/24+1/40.+1/19800
1+2+3+4……+20+21+23+24……+40的简便算法
1-2/3-1/6-1/12-1/24-1/48-……1/768
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