(1+1/2+1/3......1/100)*(1/2+1/3......1/101)-(1+1/2+1/3......1/101)=?
来源:百度知道 编辑:UC知道 时间:2024/06/18 08:29:43
(1+1/2+1/3......1/100)*(1/2+1/3......1/101)-(1+1/2+1/3......1/101)=?谢谢!
设1/2+1/3......1/101=A
(1+1/2+1/3......1/100)*(1/2+1/3......1/101)-(1+1/2+1/3......1/101)=(1+A)*A-(1+A+1/101)=A+A^2-1-A-1/101=A^2-1-1/101=(1/2+1/3......1/101)(1/2+1/3......1/101)-1-1/101
1+1/2+1/3......1/100)*(1/2+1/3......1/101)-(1+1/2+1/3......1/101)=(1+A)*A-(1+A+1/101)=A+A^2-1-A-1/101=A^2-1-1/101=(1/2+1/3......1/101)(1/2+1/3......1/101)-1-1/101
main()
{
float s,n,h,m;
int a;
s=1.0;h=1.0; 这里改一下
printf("please input a number!!!\n");
scanf("%d",&a);
for(n=2;n<=a;n++)
{
h=(-1)*h;
m=h*(1/n);
s=s+m;
}
printf("sum=%f\n",s);
}
你题目没出完,后面应该也是两个和相乘才又解的
(1+1/2+1/3+1/4)×
(1-1/2)(1-1/3)(1-1/4)(1-1/5).....(1-1/1000)
1+1/2+1+1/3+1+1/4+......+1/100=?
(1+1/2)(1+1/2^2)(1+1/2^4)(1+1/2^8)
(1-1/2^2)*(1-1/3^2)*(1-1/4^2).......(1-1/100^2)
1/1+1/2+1/3+1/4+。。。。+1/N 是多少
1/1+1/2+1/3+1/4+......1/2002=?
1-1/2+1/3-1/4+........1/99-1/100
求和Sn=1+(1+1/2)+(1+1/2+1/4)+....+[1+1/2+1/4.....+1/2^(n-1)]
数列 1+(1+1/2)+(1+1/2+1/4)+..............=?