jjdf.net
当前位置:首页 >> 数学题,1/100,1/7,1,1,4,() >>

数学题,1/100,1/7,1,1,4,()

1/1*4+1/4*7+1/7*10+…+1/97*100 =1/3x(1-1/4)+1/3x(1/4-1/7)+..+1/3x(1/97-1/100) =1/3x(1-1/100) =33/100

:假设它有一个极限(设为A)则有此式的前n项之和为A,也就是说{1/2+···+1/n=A 1/2+···+1/n+···=A 而1/n以后的项之和要等于0,我们取1/(n+1) +···+ 1/2(n+1),共有(n+1)项,而且每一项都小于其前一项, 故:1/(n+1) +···+ 1/2(n+1)< (n+1)...

没有,这是调和数列, 很早就有数学家研究,比如中世纪后期的数学家Oresme在1360年就证明了这个级数是发散的。他的方法很简单: 1 +1/2+1/3 +1/4 + 1/5+ 1/6+1/7+1/8 +... 1/2+1/2+(1/4+1/4)+(1/8+1/8+1/8+1/8)+... 注意后一个级数每一项对应的...

(1-1/100×70)÷2/7 =(1-7/10)÷2/7 =3/10÷2/7 =21/20 7/12+1/2+1/4 =7/12+6/12+3/12 =4/3

都给乘个3。则3/4=1-1/4、3/(4×7)=1/4-1/7依次类推,中间的约掉了,最后剩1-1/100=99/100,再除以开始给乘的3,的33/100

#include using std::cout; double f(const int &n){ if(n/2)return 1.0/n; else return -1.0/n; } int main() { int n(10); double res(0.0); while(n) { res+=f(n); --n; } cout

解: 1*1/4+4*1/7+7*1/10+...+97*1/100 =1/3*(1-1/4)+1/3*(1/4-1/7)+1/3*(1/7-1/10)+……+1/3*(1/97-1/100) =1/3*[(1-1/4)+(1/4-1/7)+(1/7-1/10)+……+(1/97-1/100)] =1/3*(1-1/100) =1/3*99/100 =33/100

=(1/4-1/5)+(1/5-1/6)+(1/6-1/7)+……+(1/99-1/100) =1/4-1/100 =6/25

1/2×3+1/3×4+1/4×5+1/5×6+1/6×7+……+1/99×100 =1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+……1/99-1/100 =1/2-1/100 =49/100

1.7/2=3.5,不三不四 2.1×1=1,一成不变 3.7/8 七上八下 4.15钟=一刻,千金一刻

网站首页 | 网站地图
All rights reserved Powered by www.jjdf.net
copyright ©right 2010-2021。
内容来自网络,如有侵犯请联系客服。zhit325@qq.com