(1-1/2^2)*(1-1/3^2)*(1-1/4^2)*···*(1-1/2011^2)
1、(1+3+5+···+2009+2011)-(2+4+6+···+2010+2012)
(1+1/2)(1-1/3)(1+1/4)(1-1/5)··········(1-1/2005)(1+1/2006)=_
(-1)+2+(-3)+4+······+(-2009)+2010+(-2011)=__________________
求证;1+1/2+1/3+······+1/k≧(2k)/(k+1)
计算 1-2+3-4+5-6+····+(-1)^n+1*n
1/2+(1/3+2/3)+(1/4+2/4+3/4)+···+(1/10+2/10+···9/10)
计算(+1)+(-2)+(+3)+(-4)+······+(-100)+(+101)+(-102)
2010·(1-1/2)·(1-1/3)·(1-1/4)·…·(1-1/2010)
(1)2010×(1-1/2)×(1-1/3)×(1-1/4)×·········×(1-1/2009)×(1-1/20
计算:1/2+(1/3+2/3)+(1/4+2/4+3/4)+···+(1/40+2/40+···+39/40)
根据1+2+3+4+······+n=n(n+1)/2 计算2+4+6+····+200
-1*1/2+(-1/2*1/3)+(-1/3*1/4)+···+(-1/2010*1/2011)