koExercise 5.2 Summing 100 data values

Exercise 5-2. Define an array, data, with 100 elements of type double. Write a loop that
will store the following sequence of values in corresponding elements of the array:

1/(2*3*4) 1/(4*5*6) 1/(6*7*8) ... up to 1/(200*201*202)

Write another loop that will calculate the following:

data[0] - data[1] + data[2] - data[3] + ... -data[99]

Multiply the result of this by 4.0, add 3.0, and output the final result. Do you recognize the
value you get?

//Exercise 5.2 Summing 100 data values
#include <stdio.h>

int main(void)
{
  double data[100];                    // Stores data values
  double sum = 0.0;                    // Stores sum of terms
  double sign = 1.0;                   // Sign - flips between +1.0 and -1.0
  size_t i = 0;

  int j = 0;
  for( i = 0 ; i < sizeof(data)/sizeof(double) ; ++i)// sizeof （数组名字）计算了这个数字的大小？
  {
    j = 2*(i + 1);
    data[i] = 1.0/(j * (j + 1) * (j + 2));
    sum += sign*data[i];
    sign = -sign;
 }

   // Output the result
  printf("The result is %.4lf\n", 4.0*sum + 3.0);
  printf("The result is an approximation of pi, isn't that interesting?\n");
    return 0;
    }