Sum of digitsA function, sumN, is defined that takes an integer n as argument and returns the sum of the integers from 1 through n. Write a C program to determine value of the expression shown below? sumN(3456) - sumN(3455)Note: the arguments of function sumN() can be obtained from user at runtimeTestcases:Input:18041604Output:2Input:34563455Output:1
Question
Sum of digitsA function, sumN, is defined that takes an integer n as argument and returns the sum of the integers from 1 through n. Write a C program to determine value of the expression shown below? sumN(3456) - sumN(3455)Note: the arguments of function sumN() can be obtained from user at runtimeTestcases:Input:18041604Output:2Input:34563455Output:1
Solution
Here is a simple C program that calculates the sum of integers from 1 through n and then determines the value of the expression sumN(n) - sumN(n-1).
#include <stdio.h>
// Function to calculate sum of integers from 1 to n
int sumN(int n) {
int sum = 0;
for(int i = 1; i <= n; i++) {
sum += i;
}
return sum;
}
int main() {
int n1, n2;
printf("Enter two numbers: ");
scanf("%d %d", &n1, &n2);
int result = sumN(n1) - sumN(n2);
printf("The result is: %d\n", result);
return 0;
}
This program first defines a function sumN that calculates the sum of integers from 1 to n. It then reads two numbers from the user, calculates the sum of integers from 1 to each number using the sumN function, and subtracts the two sums to get the result.
However, since the question asks for sumN(3456) - sumN(3455), the result will always be 3456 because sumN(n) - sumN(n-1) = n. This is because sumN(n) includes n in its sum, while sumN(n-1) does not. Therefore, when you subtract sumN(n-1) from sumN(n), you are left with n.
Similar Questions
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