求S=1+1/(1+2)+1/(1+2+3)+……+1/(1+2+3+4+……+n)=?数列那一章的
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求S=1+1/(1+2)+1/(1+2+3)+……+1/(1+2+3+4+……+n)=?数列那一章的
求S=1+1/(1+2)+1/(1+2+3)+……+1/(1+2+3+4+……+n)=?数列那一章的
求S=1+1/(1+2)+1/(1+2+3)+……+1/(1+2+3+4+……+n)=?数列那一章的
先求分母的表达式,1+2+3+.+n=(1+n)n/2
1/(1+2+3+...+n)=2/n(1+n)=2[1/n-1/(n+1)]
然后S就马上可以算出来了
1+1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+……+1/(1+2+3……+n)
= 1+1/[(1+2)×2÷2]+1/[(1+3)×3÷2]+……+1/[(1+n)×n÷2]——①
= 2/2+2/(1+2)×2+2/(1+3)×3+……+2/(1+n)×n——②
= 2×[1/2+1/2-1/3+1/3-1/4+……+1/n-1/(1+n)]——③
全部展开
1+1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+……+1/(1+2+3……+n)
= 1+1/[(1+2)×2÷2]+1/[(1+3)×3÷2]+……+1/[(1+n)×n÷2]——①
= 2/2+2/(1+2)×2+2/(1+3)×3+……+2/(1+n)×n——②
= 2×[1/2+1/2-1/3+1/3-1/4+……+1/n-1/(1+n)]——③
= 2×[1-1/(1+n)]
= 2×[n/(1+n)]
= 2n/(1+n)
注释:
①把分母等差数列写成简便形式
②分子和分母同时乘以2
③把分子2提出来做公因数
收起
1/(1+2+3+4+……+n)
= 2/[ n(n+1) ]
= 2[1/n - 1/(n+1) ]
因此相加得2(1/1 - 1/(n+1))
= 2n/(n+1)
#include
int main()
{
double s=0;
int n,t=0;
printf("请输入n\n");
scanf("%d",&n);
int i;
for(i=1;i<=n;i++)
{
t+=i;
s+=1.0/t;
}
printf("结果为:%f",s);
return 0;
}
求采纳!