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(1/2+1/3+...1/2015)(1+1/2+1/3+...+1/2014)%(1+1/...

最后一个括号里面应该是(1/2)+(1/3)+……+(1/2014)吧! 设1+(1/2)+(1/3)+……+(1/2015)=a 那么,原式=(a-1)×[a-(1/2015)]-a×[a-1-(1/2015)] =a(a-1)-a×(1/2015)+(1/2015)-a(a-1)+a×(1/2015) =1/2015

答: 第n项分数的分母=(1+2+3+...+n)=(n+1)n/2 第n项分数=2/[n(n+1)]=2/n-2/(n+1) 原式 =2*[1-1/2+1/2-1/3+.....+1/2013-1/2014] =2*(1-1/2014) =2013/1007

分析:算式的第一部分,先用乘法分配律展开,就容易做了: 解:原式=(1/2+1/3+...+1/2000)+ (1/2+1/3+...+1/2000)×(1/2+...+1/1999)-(1+1/2+...+1/2000)*(1/2+1/3+...+1/1999) =(1/2+1/3+...+1/2000)+(1/2+...+1/1999)×[(1/2+1/3+...+1/2000)-...

设x=1/2+1/3+...+1/2004 y=1/2+...+1/2003 (1/2+1/3+...+1/2004)(1+1/2+...+1/2003)-(1+1/2+..+1/2004)(1/2+1/3+..+1/2003) =x(1+y)-(1+x)y =x+xy-y-xy =x-y =1/2004

令a=1/2+1/3+...+1/2003 原式=a(1+a+1/2004)-(1+a)(a+1/2004) =a(1+a)+a/2004-a(1+a)-(1+a)/2004 =a/2004-(1+a)/2004 =(a-1-a)/2004 =-1/2004

直接套用公式n*1/(n+1)=1/n-1/(n+1),每项拆分后的后一项与下一项的前项消去了.如1/(1*2)+1/(2*3)=1-1/2+1/2-1/3=2/3.所以结果为1-1/2014=2013/2014

利用“欧拉公式” 1+1/2+1/3+……+1/n =ln(n)+C,(C为欧拉常数) 具体证明看下面的链接 欧拉常数近似值约为0.57721566490153286060651209 这道题用数列的方法是算不出来的 Sn=1+1/2+1/3+…+1/n >ln(1+1)+ln(1+1/2)+ln(1+1/3)+…+ln(1+1/n) =ln2+ln(3/2)...

没错吧,但是float是有误差的3+3可能不是6等等, 另外float是不能求于的

考察一般项第n项: [1/(n+1)]/[(1+1/2)(1+1/3)...(1+1/(n+1)] =[1/(n+1)]/[(3/2)(4/3)...(n+2)/(n+1)] =[1/(n+1)]/{[3×4×...×(n+2)]/[2×3×...×(n+1)]} =[1/(n+1)]/[(n+2)/2] =2/[(n+1)(n+2)] =2[1/(n+1)-1/(n+2)] (1/2)/(1+1/2)+(1/3)/[(1+1/2)(...

1+2+3+...+n=n(n+1)/2 1/(1+2+3+...+n)=2/n(n+1)=2[1/n-1/(n+1)] 1+1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+......+1/(1+2+3+...+100) =2[(1-1/2)+(1/2-1/3)+(1/3-1/4)+......+(1/100-1/101)] =2(1-1/101) =200/101

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