C programming.. for loop?

Most contributions are either over complicated and or miss the requirements.

Douzo has provided a good solution, except that his loop will be not evaluate for i == 20

The conditional test needs to be i <= 20

int i, sum;
for (i=2,sum=i;i<20;i+=2,sum+=i)
    printf("%d + ", i);
printf("%d = %d", i, sum);

n=2
str = n.tostring

for i=1 to 10
n=n+2
str = str & " + " & n.tostring
next
str = str & " = 110"

console.writeln(str)

PS: Not C code, but you get the general idea

Sorry. You should not develop such an impressions. It is an individuals preference to use i, j as loop control variables. It all depends upon the loop structure. for(i=0;i<n-1;i++) for(j=0;j<n-1-i;j++) Here, i loop runs for n-1 times while j loop runs for n-1-i times. That is, when i is 0, inner loop runs for n-1 times, when i is 1 it runs for n-2 times .....when i value is n-2 inner loop runs for 1 time

That is a straight forward one
#include <stdio.h>
void main (void)
{int loop, sum=0;
for (loop =2; loop >=20; loop+=2)
{ sum+=loop;
printf ("%d + ",loop);
}
printf (" = %d \n",sum);
return;
}

I don't know what a comma operator is.
do you mean common?

this for loop will display that result to your screen:

for (int i = 0; i < 1; ++i)
{
cout << "2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 18 + 20 = 110" << endl;
}

just remember to #include <iostream>

or alternatively you could do:

int sum = 0; int i;
for (i = 2; i < 20; i = i + 2)
{
sum += i;
cout << i << " + ";
}
cout << i << " = " << (sum += i) << endl;

int index;
int next;
int sum = 0;

for (index = 0, next = 0;
index < 10;
next += 2, sum += next, index++, printf("%d", next) )
{
if( index > 0 && index < 10)
{
printf(" + ");
}

}

printf(" = %d\n",sum);



I think this is what she needs, uses the comma operator and calculates the sum in the loop, Though Douzo Yoroshiku's method is better.
This kinda thing is rarely ever done...

include <stdio.h>

int main()
{
const int LIMIT_VALUE = 10;

int sum = 0;
for ( int loopNumber=1; loopNumber < LIMIT_VALUE+1 ; loopNumber++ )
{
int currentNumber = 2*loopNumber;
printf("%d", currentNumber);
sum = sum + currentNumber;

if (loopNumber < LIMIT_VALUE)
{
printf(" + ");
}
else
{
printf(" = ");
}
}
printf("%d", sum);

/* wait on a keypress - prevents the console disappearing on termination*/
int ch;
do
{
ch = fgetc ( stdin );
}
while ( ch != EOF && ch != '\n' );

return 0;
}

include<studio.h>
#include<conio.h>
void main()
int i=0,j=0,k=0;
for (i>=0;i>=10;i++)
{
j=j+2;
k=k+j;
}
getch();
}
if you are satisfy with my answar than pls give me reply.

include<stdio.h>
#include<conio.h>
void main()
{
int sum=0,I;
for(i=2;i<=20;i=i+2)
sum=sum+i;
printf("%d",sum);
getch();
}