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(1%1/2%1/3%1/4%1/5)(1/2+1/3+1/4+1/5+1/6)%(...

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

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+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

原式=(1-1/2-1/3-1/4-1/5)*(1/2+1/3+1/4+1/5)+1/6*(1-1/2-1/3-1/4-1/5) -(1-1/2-1/3-1/4-1/5)*(1/2+1/3+1/4+1/5)+1/6*(1/2+1/3+1/4+1/5) =1/6*(1-1/2-1/3-1/4-1/5+1/2+1/3+1/4+1/5) =1/6*1 =1/6

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...

原式=(1+1/2+1/3+1/4)×(1/2+1/3+1/4+1/5)-(1+1/2+1/3+1/4)×(1/2+1/3+1/4)-(1/5)×(1/2+1/3+1/4)=(1+1/2+1/3+1/4)×(1/5)-(1/5)×(1/2+1/3+1/4)=(1/5)×1=1/5 原式=1/2+(1/3+2/3)+(1/4+2/4+3/4)+……+(1/10+2/10+3/10+…+9/10)=1/2+1+3/2+2+5/2+3+7/2+4+...

1/3+2/3=1 1/4+2/4+3/4=2/4+1=1/2+1 1/5+2/5+3/5+4/5=(1/5+4/5)+(2/5+3/5)=1+1 1/6+2/6+3/6+4/6+5/6=(1/6+5/6)+(2/6+4/6)+3/6=1+1+1/2 应该能看出规律了吧 那么7为分母的结果为1+1+1 8为分母的结果为1+1+1+1/2 9的为1+1+1+1 10的为1+1+1+1+1/2 ...

当n=>∞时 1-1/2+1/3-1/4……+1/2n =1+1/2+1/3+1/4……+1/2n-2(1/2+1/4+……+1/2n) =1/(n+1)+1/(n+2)+……1/2n =1/n(1/(1+1/n)+1/(1+2/n)+……+1/(1+n/n) =1/(1+x)[从0积到1]=ln2

2004能被2,3,4,6,12整除,所以可以不考虑1/2,1/3,1/4,1/6,1/12 2004除以5,8,10是有限小数,所以也可以不考虑1/5,1/8,1/10 2004/7=286.285714285714,是一个6位的循环,小数点后第2004位,2005位是42 2004/9=222,666是一个1位的循环...

#include int main () { int i,n,m=-1,k=0; double j,sum=0.0; scanf("%d",&n); for(i=1;i

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