C programming adding natural numbers

Question

I need my program to run and tell me the sum of the natural number entered also it needs to say along with the sum total i need it to show the sum of odd and even integers. This is what I have so far and it won't run correctly in C.

#include <stdio.h>
int main (void)
{
    int n, i, sum = 0;
    int sum1 = 0;
    int sum2 = 0;
    printf("enter a number and I will tell you the numbers sums.");
    scanf("%d", &n);

    for(i=1; i<= n; ++n)
    {
        sum2 = sum2 + n;
    }
    for(i=2; i<= n; ++n)
    {
        sum1 = sum1 + n;
    }
    for(i=1; i<= n; ++n)
    {
        sum += i;
    }
    printf("sum of integers is %d" ,sum);
    printf("sum of odd integers is %d" ,sum1);
    printf("sum of even integers is %d" ,sum2);

    return 0;
}

needs minimal reproducible example and a proper problem statement. — Antti Haapala
– Antti Haapala, Commented Feb 9, 2017 at 0:59
i.e. "Describe the problem. "It doesn't work" is not a problem statement. Tell us what the expected behavior should be. Tell us what the exact wording of the error message is, and which line of code is producing it. Put a brief summary of the problem in the title of your question." — Antti Haapala
– Antti Haapala, Commented Feb 9, 2017 at 1:00

Rishi · Accepted Answer · 2017-02-09 14:26:51Z

1

In your loops to count odd and even, you need to increment by 2 in the loop, not one. Instead of ++i, use i += 2:

for (i = 2; i <= n; i += 2)

And it should be i in the increment, not n. You're changing the value of your final variable. Further, unless I'm misunderstanding what you're trying to do, you should be adding i to your sums, not n.

edited Feb 9, 2017 at 14:26

Rishi

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answered Feb 9, 2017 at 0:42

goug

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Comments

ChuckCottrill · Accepted Answer · 2017-02-10 00:08:24Z

1

The sum of the first N natural numbers is a well-known formula,

sum(range(1,N)) == N*(N+1)/2

Read this exposition, and then see if you can derive a formula for the sum of even or odd, http://mathandmultimedia.com/2010/09/15/sum-first-n-positive-integers/

(hint: the sum of every other number would be about 1/2)

answered Feb 10, 2017 at 0:08

ChuckCottrill

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Comments

user7528903 · Accepted Answer · 2017-02-10 00:47:01Z

0

You only require a one for loop here's how

sum = sum1= sum2 =0;
for(i=0; i<= n; i++)
{
sum = sum +i;
if(i %2 == 0)
sum2 += i;
else
sum1 +=i;
}

use if statement for filtering numbers and don't forget to initialise all sum var to 0

answered Feb 10, 2017 at 0:47

user7528903

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