for loop

[w0ot]

Would someone kindly show me the code of using a for loop to obtain and display the sum of odd integer in the range of 11 to 121 ?

thanks !

[Danielsson]

Just a hint: odd integers meet this condition: x % 2 == 1

[Maimaje Bello Abdullahi]

int sum=0;

for(int i=11; i<=121; i++)
{
     if(i%2 == 1)
      {
          sum = sum + i;
       }
}
cout<<sum<<endl;

Hope this will help you

[sammy34]

It would be more efficient to increment i by 2 each time:

int sum=0;

for(int i=11; i<=121; (i + 2))
{
          sum = sum + i;
}
cout<<sum<<endl;

This way you're only iterating over half the values, and saving a check each time.

(Note: You can probably use a variation of Euler's counting formula to solve this without using a for loop)

[sammy34]

Oops I made a mistake. The update condition in the for loop above should read (i = i+2).

[sammy34]

Actually the formula for the sum of the first N odd numbers (1, 3, 5, ...(2N-1)) is simply N^2. All you have to do then is take into account the offset. It works out that the sum of N odd numbers starting at odd number x is N(N + x - 1).

So, from 11 to 121 inclusive, there are 56 odd values if I'm not mistaken. You're starting at 11, so you could just write:

cout << 56*(56 + 11 - 1)

Which gives 3696, which I believe is correct. But I guess the point of the exercise was to learn about for loops, not counting sequences hehe :).