Question

find the sum of all numbers upto n

Answer 1

#include<stdio.h>

int main (){

int i,n,sum=0;

printf("Please enter a number:");

scanf("%d",&n);

for(i=1;i<=n;i++){

sum=sum+i;

}

printf("Sum of all number to n is : %d",sum);

return 0;

}

Answer 2

//i am doing by using c language

#include<stdio.h>

void main()

 {

  int i,n,sum;

printf("enter value of n");

scanf("%d",&n);

for(i=1;i<=n;i++)

{

sum+=i;

}

printf("sum is %d",sum);

}

Answer 3

n=int(input("enter the last number ypu want:"))
sum=0
for i in range(0,n):
    sum+=i
print(sum)

Answer 4

uint32_t sum(uint32_t n) {

    return n*(n+1)/2; // thank you gauss! 

}

Answer 5

// same result without loop for

#include <stdio.h>

int main()
{
    int i,n,sum=0,sum1=0;
    
    printf("Please enter a number:");
    scanf("%d",&n);
    sum=(n+1)*n/2;
    printf("Sum of all number to n is : %d",sum);

    return 0;
}

Answer 6

I used a simple formula from Arithmetic  Progression

#include<stdio.h>

void main()

 {

  int n,sum;
  printf("ENTER VALUE OF n:");
  scanf("%d",&n);
  sum=0.5*n*(1+n);
  printf("\n sum=%d",sum);
  
}

Answer 7

#include <stdio.h>

int main

{

int a,i,c=0;

printf("ENTER THE NUMBER");

scanf("%d",&a0;

for(i=0;i<=a;i++)

c=c+i;

printf("the required sum%d",c);

return 0;

}

Answer 8

#include<stdio.h>
void main()
{
int i,n ,sum=0;
printf("Enter the number:");
scanf("%d",&n);
for(i=0;i<=n;i++)
{
sum=sum+i;
}
printf("The sum of numbers upto 'n' is %d",sum);
}