1+1/(1+2)+1/(1+2+3)+·····+1/(1+2+3+·····+n) 求和
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1+1/(1+2)+1/(1+2+3)+·····+1/(1+2+3+·····+n) 求和
1+1/(1+2)+1/(1+2+3)+·····+1/(1+2+3+·····+n) 求和
1+1/(1+2)+1/(1+2+3)+·····+1/(1+2+3+·····+n) 求和
1+2+3+…+n=n(n+1)/2 得1/(1+2+3+…+n)=2/n(n+1)=2/(1/n - 1/(n+1)) 故 1+1/(1+2)+1/(1+2+3)+……+1/(1+2+3+…+n) =2/(1/1 - 1/(1+1)) + 2/(1/2 - 1/(2+1))+...+2/(1/n - 1/(n+1)) =2/(1/1 - 1/(n+1)) =2n/(n+1)
(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
巧算(奥数题)问:(1+1/2)×(1-1/2)×(1+1/3)×(1-1/3)×...(1+1/99)×(1-1/99)
(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)
2000*(1-1/2)*(1-1/3)*.*(1-1/1999)*(1-1/2000)
(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?