1+1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+……+1/(1+2+3+4……+10)等于几?
来源:学生作业帮助网 编辑:作业帮 时间:2024/11/13 04:35:22
1+1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+……+1/(1+2+3+4……+10)等于几?
1+1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+……+1/(1+2+3+4……+10)等于几?
1+1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+……+1/(1+2+3+4……+10)等于几?
1+1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+……+1/(1+2+3+4……+10)
=1/ (1*2 /2) +1/ (2*3 /2) +1/ (3*4 /2) +1/ (4*5 /2)+……+1/ (10*11 /2)
=2/ 1*2 +2/ 2*3 +2/ 3*4 +2/ 4*5 +……+ 2/ 10*11
=2 * (1/ 1*2 +1/ 2*3 +1/ 3*4 +1/ 4*5+……+1/ 10*11)
=2 * (1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+……+1/10-1/11)
=2 * (1-1/11)
=2 * 10/11
=20/11
就是自然数列求和及拆项法的组合.
20/11
(1/2+1/3+...+1/2004)(1+1/2+1/3+...+1/2003)-(1+1/2+1/3+...+1/2004)(1/2+1/3+...+1/2003)
200*(1-1/2)*(1-1/3)*(1-1/4)*.*(1-1/100)
(1-1/2^2)(1-1/3^2)K(1-1/10^2)
(1-1/2^2)(1-1/3^2)K(1-1/10^2)
(1-2/1)*(1-3/1)*(1-4/1)*.*(1-2007/1)*(1-2008/1)
(1/2014-1)(1/2013-1)(1-2012-1)...(1/3-1)(1/2-1)
2000*(1-1/2*)*(1-1/3)*...*(1-1999)*(1-1/2000)
计算:(-1)-[1-(1-1/2*1/3)]*6
计算!(-1)-[1-(1-1/2*1/3)]*6
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(1/2+1/3+...+1/2007)*(1+1/2+1/3+...+1/2006)-(1+1/2+1/3+...+/2007)*(1/2+1/3+...+1/2006)
计算:(1/2+1/3+...+1/2011)*(1+1/2+1/3+...+1/2010)-(1+1/2+1/3+...+1/2011)*(1/2+1/3+...+1/2010)
计算(1-1/2-1/3-...-1/2010)*(1/2+1/3+..+1/2011)-(1-1/2-1/3-...-1/2011)*(1/2+1/3+...+1/2010)
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(1-1/2004)(1-2003)(1-2002)…(1-1/3)(1-1/2)
(1-1/2)(1-1/3)(1-1/4).(1-1/100)
(1/1+2)+(1/1+2+3)+…+(1/1+2+3…+2000)
(1+1/2)*(1-1/2)*(1+1/3)*(1-1/3)*...(1+1/2006)*(1-1/2006)怎么简算呀?