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1/1*2+1/2*3+1/3*4+1/4*5+1/5*6+1/6*7 简便运算怎...

拆分每个分数。记住等式: 1/(x(x+1))=1/x-1/(x+1) 所以,原等式就可以简化为1/1-1/10=0.9

解法一: 1×2+2×3+3×4+...+n(n+1) =⅓×[1×2×3-0×1×2+2×3×4-1×2×3+3×4×5-2×3×4+...+n(n+1)(n+2)-(n-1)n(n+1)] =⅓n(n+1)(n+2) 解法二: 考察一般项第k项,k(k+1)=k²+k 1×2+2×3+3×4+...+n(n+1) =(1²+2²+3²+...+n...

1/1*2+2/1*2*3+3/1*2*3*4+4/1*2*3*4*5+5/1*2*3*4*5*6+6/1*2*3*4*5*6*7 =1-1/1*2+1/1*2-1/1*2*3+1/1*2*3-1/1*2*3*4+1/1*2*3*4-1/1*2*3*4*5+1/1*2*3*4*5-1/1*2*3*4*5*6+61//1*2*3*4*5*6-1/1*2*3*4*5*6*7 =1-1/1*2*3*4*5*6*7 =1-1/5040 =5039/5040...

1/1*2+1/2*3+1/3*4+1/4*5+1/5*6+1/6*7 =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

1/1*2+1/2*3+1/3*4+1/4*5+1/5*6+……1/599*600 =1-1/2+1/2-1/3+1/3-1/4+……+1/599-1/600 =1-1/600 =599/600

从题目看应先输入项数n(正整数),再用for循环求结果较方便;计数变量从1开始依次增1,奇数取其倒数为正、偶数取其倒数为负求各项之浮点和即可完成。代码如下: #include "stdio.h"int main(int argc,char *argv[]){double s;int i,n;printf("Inp...

=1-1/2+1/2-1/3+1/3-1/4+...+1/2016-1/2017 =1-1/2017 =2016/2017

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

这道题目用错位相减法来做。 错位相减法是一种常用的数列求和方法,应用于等比数列与等差数列相乘的形式。 形如An=BnCn,其中Bn为等差数列,Cn为等比数列;分别列出Sn,再把所有式子同时乘以等比数列的公比,即kSn;然后错一位,两式相减即可。 ...

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