①1/2+1/2*3+1/3*4+…+1/2006*2007=②[(1/2007)-1]*[(1/2006)-1]*[(1/2005)-1]…*[(1/2)-1]=
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①1/2+1/2*3+1/3*4+…+1/2006*2007=②[(1/2007)-1]*[(1/2006)-1]*[(1/2005)-1]…*[(1/2)-1]=①1/2+1/2*3+1/3*4+…
①1/2+1/2*3+1/3*4+…+1/2006*2007=②[(1/2007)-1]*[(1/2006)-1]*[(1/2005)-1]…*[(1/2)-1]=
①1/2+1/2*3+1/3*4+…+1/2006*2007=
②[(1/2007)-1]*[(1/2006)-1]*[(1/2005)-1]…*[(1/2)-1]=
①1/2+1/2*3+1/3*4+…+1/2006*2007=②[(1/2007)-1]*[(1/2006)-1]*[(1/2005)-1]…*[(1/2)-1]=
1.1/2+1/2*3+1/3*4+…+1/2006*2007
=1-1/2+1/2-1/3+1/3-1/4+……+1/2006-1/2007
=1-1/2007=2006/2007
2.[(1/2007)-1]*[(1/2006)-1]*[(1/2005)-1]…*[(1/2)-1]
=[1/2-1][1/3-1][1/4-1]……[1/2006-1][1/2007-1]
=(-1/2)(-2/3)(-3/4)……(-2005/2006)(-2006/2007)
=1/2007
(1-1/2)+(1/2-1/3)+...+(1/2006-1/2007)=1-1/2007=2006/2007
(-2006/2007)*(-2005/2006)*...*(-2/3)*(-1/2)=1/2007(注:一共是2006个负数,他们的积为正)
1/(1-1/2)/(1-1/3)/(1-1/4)/……/(1-1/2012)
2(3+1)(3^2+1)(3^4+1)……(3^32+1)+1
|1/2-1|+|1/3-1/2|+|1/4+1/3|+…+|1/30-1/29|
1/1+2+1/1+2+3+1/1+2+3+4+…+1/1+2+3+…+50
求证:1/2^3 +1/3^3 +1/4^3 +……+1/(n+1)^3
求证:1/2^3 +1/3^3 +1/4^3 +……+1/(n+1)^3
数学无敌者进求和:1/2+1/3+1/4+1/5+……我的证明:所求为S,1/2*S=1/4+1/6+1/8+……,①S-①:1/2*S=1/2+1/3+1/5+1/7+1/9+……,明显①≠②
求和:1/1×2+1/2×3+1/3×4+……+1/n(n+1)
(1-1/2)×(1-1/3)×(1-1/4)……×(1-1/2008)计算方法
200×(1-1/2)×(1-1/3)×(1-1/4)×……×(1-1/100)=?
计算:(1-1/2)*(1-1/3)*(1-1/4)*……*(1-1/10).
(1/2+1/3+1//4+…+1/2005)(1+1/2+1/3+1/4+…+1/2004)-(1+1/2+1/3+1/4+…+1/2005)(1/2+1/3+1/4+…+1/2004计算1/2+1/3+1//4+…+1/2005)(1+1/2+1/3+1/4+…+1/2004)-(1+1/2+1/3+1/4+…+1/2005)(1/2+1/3+1/4+…+1/2004)
(1/2+1/3+1/4+……+1/2013)(1+1/2+1/3+1/4+……+1/2012)-(1+1/2+1/3+……+1/2013)(1/2+……+1/2012)
计算(1+3)(1+3^2)(1+3^4)……(1+3^64)+1
①1/2+1/2*3+1/3*4+…+1/2006*2007=②[(1/2007)-1]*[(1/2006)-1]*[(1/2005)-1]…*[(1/2)-1]=
计算(1+3)(1+3^2)(1+3^4)……(1+3^2n)
(3+1)(3^2+1)(3^3+1)(3^4+1)…(3^32+1)步骤
(1/1+2)+(1/1+2+3)+(1/1+2+3+4)+………+(1/1+2+3+………+100)