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(1+1/2+1/3+1/4)x(1/2+1/3+1/4+1/5)-(1+1/2+1...

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

随后很长一段时间,人们无法使用公式去逼近调和级数,直到无穷级数理论逐步成熟。1665年牛顿在他的著名著作《流数法》中推导出第一个幂级数: ln(1+x) = x - x^2/2 + x^3/3 - ... Euler(欧拉)在1734年,利用Newton的成果,首先获得了调和级数有...

//#include "stdafx.h"//vc++6.0加上这一行.#include "stdio.h"int main(void){ int m,i; double s; while(1){ printf("Enter m(m>0)...\nm="); scanf("%d",&m); if(m>0) break; printf("Error, redo: "); } for(s=0,i=1;i

解析: 1+1/2+1/6+...+1/(99×100) =1+(1-1/2)+(1/2-1/3)+...+(1/99-1/100) =2-1/100 =199/100

这道题要求计算能力很强 3/x+1/(x-1)+4/(x-2)+4/(x-3)+1/(x-4)+3/(x-5)=0 [3/x+3/(x-5)]+[1/(x-1)+1/(x-4)]+[4/(x-2)+4/(x-3)]=0 (6x-15)/(x^2-5x)+(2x-5)/(x^2-5x+4)+(8x-20)/(x^2-5x+6)=0 (2x-5)[3/(x^2-5x)+1/(x^2-5x+4)+4/(x^2-5x+6)]=0 所...

s+=1.0/(i+1);

原式=1/2+1/6+1/12+1/20+1/30+1/42 =(1-1/2)+(1/2-1/3)+(1/3-1/4)+(1/4-1/5)+(1/5-1/6)+(1/6-1/7) =1-1/7 =6/7 公式: 1/a乘1/(a+1)=1/a-1/(a+1)其中a是正整数

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

1/(1×3)+1/(2×4)+1/(3+5)……+1/(2016x2018) = 1/2×{1-1/3+1/2-1/4+1/3-1/5+.....1/2016-1/2018} = 1/2 ×{1-1/3+1/3-1/5+.....1/2015-1/2017} + 1/2×{1/2-1/4+1/4-1/6+.....1/2016-1/2018} = 1/2 ×{1-1/2017} + 1/2×{1/2-1/2018} = 1/2 ×{2015/201...

24×(1/2+1/3-1/4) = 24x 1/2 + 24 x 1/3 - 24 x 1/4 =12 + 8 -6 =14

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