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(1+1/2+1/3+1/4)*(1/2+1/3+1/4+1/5)%(1+1/2+1/3+1/4+

设(1-1/2-1/3-1/4-1/5)为a,(1/2+1/3+1/4+1/5)为b,代入得转化为a(b+1/6)-(a-1/6)b,得到ab+1/6a-ab+1/6b,化简为1/6(a+b),再重新代入,解得1/6(1-1/2-1/3-1/4-1/5+1/2+1/3+1/4+1/5)=1/6*1=1/6

设+1/2+1/3+1/4=x 1/2+1/3+1/4+1/5=y (1+1/2+1/3+1/4)*(1/2+1/3+1/4+1/5)-(1+1/2+1/3+1/4+1/5)*(1/2+1/3+1/4) =(1+x)y-(1+y)x =y+xy-x-xy =y-x =1/5

1/2+(1/3+2/3)+(1/4+2/4+3/4)+……(1/50+2/50+…+48/50+49/50)= 先总结一下,凡是分母是奇数的,如(1/3+2/3)=1 (1/5+2/5+3/5+4/5)=2,都是整数,且等于(奇数-1)/2 以此类推,(1/49+2/49+…+48/49)= 24 分母是偶数的,如1/2=0.5,(1/4+2/4+3/4)=1.5...

设a=1/2+1/3+1/4 b=1/2+1/3+1/4+1/5 (1+1/2+1/3+1/4)x(1/2+1/3+1/4+1/5)-(1/2+1/3+1/4)x(1+1/2+1/3+1/4+1/5) =(1+a)b-a(1+b) =b+ab-a-ab =b-a =1/5

应该是没有什么错的,估计是你用的编译器不支持在声明完变量,执行完其它语句之后,再来临时声明新变量的功能 这样改一下就行了: #include "stdio.h"int main(){ int n,i; double sum=0; //放在这儿来声明 scanf("%d",&n); for(i=1;i

(1/2+1/3+1/4+1/5)*(1/3+1/4+1/5+1/6)-(1/2+1/3+1/4+1/5+1/6)*(1/3+1/4+1/5) 设1/2+1/3+1/4+1/5=a ,1/3+1/4+1/5=b (1/2+1/3+1/4+1/5)*(1/3+1/4+1/5+1/6)-(1/2+1/3+1/4+1/5+1/6)*(1/3+1/4+1/5) =a*(b+1/6)-(a+1/6)*b =(a-b)/6 =(1/2)/6=1/12

(1/2+1/3+1/4+1/5)X(1/3+1/4+1/5+1/6-《1/3+1/4+1/5+1/6》)=0

(1+1/2+1/3+1/4)*(1/2+1/3+1/4)-(1/2+1/3+1/4+1/5)*(1/2+1/3+1/4) =(1/2+1/3+1/4)*[(1+1/2+1/3+1/4)-(1/2+1/3+1/4+1/5)] =(1/2+1/3+1/4)*(1-1/5) =(1/2+1/3+1/4)*4/5 =13/12 * 4/5 =13/15

7.187326 累死我了。。。。。 我是谁? EXCEL

当n很大时,有:1+1/2+1/3+1/4+1/5+1/6+...1/n = 0.57721566490153286060651209 + ln(n)//C++里面用log(n),pascal里面用ln(n) 0.57721566490153286060651209叫做欧拉常数 to GXQ: 假设;s(n)=1+1/2+1/3+1/4+..1/n 当 n很大时 sqrt(n+1) = sqrt(n...

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