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1/1×2+1×1/2×3+1/3×4……+1/50×51,简算。

简便算法:利用拆分法.原式=1-1/2=1/2-1/3+1/3-1/4+……+1/9-1/10 =1-1/10 =9/10拆分法: 1/[(n-1)*n]=1/(n-1)-1/n

1-1/2=(2-1)/1*2=1/1*21/2-1/3=(3-2)/2*3=1/2*31/3-1/4=(4-3)/3*4=1/3*41/4-1/5=(5-4)/4*5=1/4*5.1/39-1/40=(40-39)/39*40=1/39*40所以1/1*2+1/2*3+1/3*4+1/4*5……+1/39*40=(1-1/2)+(1/2-1/3)+(1/3-1/4)+(1/

解:用拆项法:原式=1/2*[2/(1*2*3)+2/(2*3*4)++2/(98*99*100)] =1/2*[(3-1)/(1*2*3)+(4-2)/(2*3*4)++(100-98)/(98*99*100)] =1/2*[3/(1*2*3)-1/(1*2*3)+4/(2*3*4)-2/(2*3*4)++100/(98*99*100)-98/(98*99*100)] =1/2*[1/(1*2)-1/(2*3)+1/(2*3)-1/(3*4)++1/(98*99)-1/(99*100)] =1/2*[1/2-1/9900] =4949/19800

因为1/(n*(n+1))=1/n-1/(n+1) 所以原式=(1-1/2)+(1/2-1/3)+(1/3-1/4)++(1/18-1/19)+(1/19+1/20) =1-1/20 =19/20

1/1*2+1/2*3+1/3*4+…+1/9*10=(1-1/2)+(1/2-1/3)+(1/3-1/4)+().+(1/9-1/10)=1-1/2+1/2-1/3+1/3-1/4++1/9-1/10=1-1/10=9/10 扩展资料 分数可以表述成一个除法算式:如二分之一等于1除以2.其中,1 分子等于被除数,- 分数线等于除号,2 分母等

1/1*2+1/2*3+1/3*4+1/4*5.+1/99*100=1-1/2+1/2-1/3+1/3-1/4++1/99-1/100=1-1/100=99/100

1*1/2+1/2*1/3+1/3*1/4+……1/49*1/50=1-1/2+1/2-1/3+1/3-1/4++1/49-1/50=1-1/50=49/50

1*1/2+2*1/3+3*1/4……+99*1/100=(1-1/2)+(1-1/3)+(1-1/4)+……+(1-1/100)=99-(1/2+1/3+1/4+……+1/100)=100-(1+1/2+1/3+……+1/100) 到这里如何算后面1/2加到1/100,我貌似没有什么公式可算,只能用欧拉常数来算个大概=100-ln(100)+c(C是常数0.5822,ln(100)约等于4.6051701859) 可近似等于100-5.2 等于94.8 比较恶心,也许有高人有更好的办法吧.

** * @(#)Test.java * * * @author * @version 1.00 2009/11/29 */ public class Test { public static void main (String[] args) { float i= 1,j = 2; float a ,b=0; for ( i = 1; i<50; i++,j++) { a = 1/(i*j); b += a; } System.out.println (b); } }

1/1*2+1/2*3=2/3 1/1*2+1/2*3+1/3*4=3/41/1*2+1/2*3+1/3*4+1/4*5=4/5 所以式子=19/20

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