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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...

1/1*2+1/2*3+1/3*4+1/4*5+1/5*6+···+1/99*100 =(1-1/2)+(1/2-1/3)+(1/3-1/4)...+(1/99-1/100) =1-1/2+1/2-1/3+1/3-1/4+....1/99-1/100 =1-1/100 =99/100

1+1/2+1/3+1/4+1/5+1/6脱式计算 =1+30/60+20/60+15/60+12/60+10/60 =1+50/60+27/60+10/60 =1+87/60 =2又9/20

1/1*2*3+1/2*3*4+1/3*4*5+1/4*5*6+1/5*6*7+1/6*7*8+1/7*8*9+1/8*9*10+1/9*10*11 =(1/2)[(1/1*2)-(1/2*3)]+(1/2)[(1/2*3)-(1/3*4)]+(1/2)[(1/3*4)-(1/4*5)]+...+(1/2)[(1/9*10)-(1/10*11)] =(1/2)[(1/1*2)-(1/2*3)+(1/2*3)-(1/3*4)+(1/3*4)-(1/4*...

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

可以直接学STM32 不过学好51单片机对学习STM32肯定是有邦助的, 起码一些常见外围器件都知道怎么工作的

这是调和数列,没有简便方法 定义1:自然数的倒数组成的数列,称为调和数列. 定义2:若数列{an}满足1/a(n+1)-1/an=d(n∈N*,d为常数),则称数列{an}调和数列 人们已经研究它几百年了.但是迄今为止没有能得到它的求和公式只是得到它的近似公式(当n...

(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

=(1-5/9)*5/7 =4/9*5/7 20/63 =1-5/25*21/7 =1-3/5 =2/5

1/2+1/3+1/4+1/5+1/6+、、、1/2016=(1008+1007+1006+1005+、、、+3+2+1)/2016 =5536/2016=2.746

参考代码: #includeint main(){ int sign=1; double deno=2.0,sum=1.0,term; while (deno

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