1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)+……+1/(x+1998)(x+1999),并求当x=1时,该代数式的值.
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1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)+……+1/(x+1998)(x+1999),并求当x=1时,该代数式的值.1/x(x+1)+1/(x+1)(x+2)+1/(x+2
1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)+……+1/(x+1998)(x+1999),并求当x=1时,该代数式的值.
1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)+……+1/(x+1998)(x+1999),并求当x=1时,该代数式的值.
1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)+……+1/(x+1998)(x+1999),并求当x=1时,该代数式的值.
1/x(x+1)=1/x-1/(x+1)
1/(x+1)(x+2)=1/(x+1)-1/(x+2)
1/(x+2)(x+3)-1/(x+2)-1/(x+3)
……
1/(x+1998)(x+1999)＝1/(x+1998)-1/(x+1999)
所以,
1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)+……+1/(x+1998)(x+1999)
＝1/x－1/(x+1999)
当x＝1时,
1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)+……+1/(x+1998)(x+1999)
＝1/x－1/(x+1999)
＝1-1/2000
＝1999/2000
#include
#define x 1
void main()
{
int n;
float sum;
for(n=1,sum=0;n<=1999;n++)
{
sum=sum+1.0/((x+n-1)*(x+n));
}
printf("%f",sum);
}
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