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(1-1/4)(1-1/9)(1-1/16)···(1-1/100)

根据平方差公式, 原式 = (1-1/2)(1+1/2) * (1-1/3)(1+1/3) * (1-1/4)(1+1/4) *......*(1-1/100)(1+1/100) = 1/2*3/2*2/3*4/3*3/4*5/4*......*99/100*101/100 = 1/2*101/100 = 101/200

(1-1/4)(1-1/9)(1-1/16)(1-1/25)(1-1/36)(1-1/49)(1-1/64)(1-1/81)(1-1/100) =(1-1/2)(1+1/2)(1-1/3)(1+1/3)``````(1-1/10)(1+1/10) =1/2*3/2*2/3*4/3``````*9/10*11/10 =11/20

原式 =(1-1/2)(1+1/2) * (1-1/3)(1+1/3) * ... * (1-1/10)(1+1/10) =1/2 * 2/3 * 3/4 * ... * 9/10 * 3/2 * 4/3 * 5/4 * ... * 11/10 整理,抵消 =1/2 * 11/10 =11/20

(1-1/4)*(1-1/9)*(1-1/16)*(1-1/25)...*(1-1/100) =(1+1/2)(1-1/2)*(1+1/3)(1-1/3)*...*(1+1/10)(1-1/10) =3/2*1/2*4/3*2/3*5/4*3/4*.....*11/10*9/10 =(3/2*4/3*5/4*...*11/10)*(1/2*2/3*3/4*...*9/10) =11/2*1/10 =11/20

(1-1/4) (1-1/9) (1-1/16).....(1-1/81) (1-1/100) =[1-1/2][1+1/2][1-1/3][1+1/3][1-1/4][1+1/4]...[1-1/9][1+1/9][1-1/10][1+1/10] =1/2*3/2*2/3*4/3*3/4*5/4*...*8/9*10/9*9/10*11/10 =1/2*11/10 =11/20

解:(1-1/4)(1-1/9).....(1-1/100)=(1-1/2)(1+1/2)(1-1/3)(1+1/3)...(1-1/9)(1+1/9)(1-1/10)(1+1/10)=1/2*3/2*2/3*4/3*3/4*...8/9*10/9*9/10*11/10观察有很多项约掉了,最后得:=1/2*11/10=11/20满意谢谢采纳!

答案是:9.450232 代码如下: #include #include int main(){ double result=0; int i; double lone; for(i=1;i

(1-1/4)(1-1/9)(1-1/16)...(1-1/100) =(1-1/2²)(1-1/3²)(1-1/4²)...(1-1/10²) =(1+1/2)(1-1/2)(1+1/3)(1-1/3)(1+1/4)(1-1/4)...(1+1/10)(1-1/10) =(3/2)(1/2)(4/3)(2/3)(5/4)(3/4)...(11/10)(9/10) =[(1/2)(2/3)...(9/10)][...

原式=(1-1/2)(1+1/2)(1-1/3)(1+1/3)(1-1/4)(1+1/4)......(1-1/10)(1+1/10) =1/2 x 3/2 x 2/3 x 4/3 x 3/4 x 5/4 x ...... x 9/10 x 11/10 =1/2 x 11/10 =11/20

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