This problem requires programming, calculates interlaced sequence 1-2/3+3/5-4/7+5/9-6/11+...and the first N entries.
Input format:
Enter a positive integer N in one line.
Output format:
The partial sum value is output in one line, and the result is kept to three decimal places.
Input sample:
5
Sample output:
0.917
Code:
# include <stdio.h>
# include <stdlib.h>
int main() {
int N,i,j = 1;
scanf("%d",&N);
double sum = 0.0,m = 1.0;
for (i=1;i<=N;i++) {
sum += (i / m) * j;
j *= (-1);
m += 2;
}
printf("%.3lf",sum);
return 0;
}
Submit screenshot:
Problem-solving ideas:
Look for regular problems. Everyone should look at him carefully. At the beginning, I regarded the denominator as a Fibonacci sequence, but it was always wrong. Later, I saw that the denominator is just a simple odd term! The rest of the operation is similar to the above question!