1/2+1/3+1/4......1/100的计算方法?
来源:百度知道 编辑:UC知道 时间:2024/06/16 15:25:50
写出计算过程
编程的吗?
double count;
for (int i; i <= 100; i++)
{
count = count + (1.0 / i); //此处不能为1吧,应该是1.0
}
Response.Write(count);
double count;
for (int i; i <= 100; i++)
{
count = count + (1 / i);
}
Response.Write(count);
等差数列公式,前项加后项乘以2在除以2
所以=【(1/2+1/100)*2】/2=0.51
1+1/2+1/3+.....+1/n
1+1/2+1/3+...+1/100
1-1/2+1/3-.....-1/10
(1+1/2+1/3+1/4)×
1+1/(1+2)+1/(1+2+3)+...+1/(1+2+3+...+100)
1+1/(1+2)+1/(1+2+3)+-------+1/(1+2+3+----+100)
(1/2005-1)(1/2004-1)........(1/3-1)(1/2-1)
1+1/2+1/3+.....+1/2005][1/2+1/3+.......+1/2006]-[1/2+1/3+......+1/2005][1+1/2+1/3+1/2006]
(1-1/2-1/3-...-1/2001)*(1/2+1/3+...1/2002)-(1-1/2-1/3-...1/2002)*(1/2+1/3+...+1/2001)
1/1+1/2+1/3+1/4+。。。。+1/N 是多少