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1x3x5

1x3x5x…x1991的末三位数是(125)

1/1x3x5+1/3x5x7+1/5x7x9+…+1/95x97x99 =1/4(1/(1*3)-1/(3*5)+1/(3*5)-1/(5*7)+.....+1/(95*97)-1/(97*99) = (1/4) ( 1/(1x3) - 1/(97x99) = (1/4) (1/3 - 1/9603] =800/9603

不太简便:

其实找到了规律就很简单。 1x3x5=15,末位是5, 15x7=105,末位是5, 105x9=945,末位是5, 945x11=10395,末位是5, 10395x13=135135,末位是5. 可得接下来的积末位数字都是5 所以1x3x5x7`````x1991的末位数字是5. 望提问者采纳,谢谢!

1/1x3-1/97x99=9600/28809 不知道对不对 可以分解为An=1/(2n-1)(2n+1) -1/(2n+1)(2n+3) 1/3-1/3x5+1/3x5-1/5x7.............+1/97x99=上面第一行的答案

一.1行:方形。开始。2行:平行四边形。输入:n=1;A=1。3行:方形。A=An。4行:菱形。n《99?5行:Y。6行:方形。n=n+2。(流程线折返3、4行间)。7行:N。8平行四边形。输出A。9行:方形。结束。二.程序。1行:INPUT:n=1。2行:A=1。3行:WHILE:n《99。4行:A...

Visual Basic Dim a As Integer, sum As Integersum = 1For a = 1 To 99 Step 2 sum = sum * aNext a Print "1*3*5*...*99="; sumC int a,sum;for(a=1,sum=1;a

#include void main() { int a,b,b;//打错了吧…呵呵 ^_^ b=1; a=b*c; //错c未赋值且没有定义 c=c+2; if (c>=11),end; //c中没有(,end) else return(a);//不需要返回语句 ,这样会退出函数的return;语句有退出功能 printf("a"); //打印语句中...

an=1/[(2n-1)(2n+1)(2n+3)] =(1/4) { 1/[(2n-1)(2n+1)] -1/[(2n+1)(2n+3)] } Sn =a1+a2+...+an 1/(1x3x5)+1/(3x5x7)+1/(5x7x9)+...+1/(95x97x99) =S48 = (1/4) ( 1/(1x3) - 1/(97x99)] = (1/4) (1/3 - 1/9603] =800/9603

(1+2015)×1008÷2 =2016×504 =1016064

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