loop problem?

You should write code for it. :)

Take a pen and paper, write down the steps that need to be performed, then write the same steps in C# code.

try this thing in your program
It works.

It doesn't work. Secondly, why are you trying to do this guys homework for him? Anyone with a week of programming experience in any language should be able to knock that out within 30 minutes in c#.

Let him at least put up his own attempt.

Also, don't fix your solution, at least make him do that.

A program will calculate and display the sum of all even even numbers from 2 to n, where n is a positive number inputted by the user. What should I do and what is the expression of that?

This is one way you can do it .

int sum = 0;
for ( int i = 2; i <= n; i += 2 ) {
sum += i;
}

Well, the problem with your original code (from what i remember, i only glanced at it when i noticed the problem) was that you coded the loop like this:

for(int i=0; i <= n; n++)
DoStuff();

So unless i misread it, it couldn't have worked. Maybe you transcribed it wrong, or maybe i misread. Either way, if that's what you did write, it was a mistake.

i didnot write the full code because it was his homework problem , i just wanted to give him hint. I will send the complete code to by private message,
This problem is not new this is asked many times in c++, java etc.
klamath23

public uint Calculate(uint n)
{
if (n % 2 == 1)
n -= 1;

return Convert.ToUInt32(n*(n/4 + 0.5));
}

Suppose n=6. The final answer should be 6 + 4 + 2 = 12.

In your code, if n=6, the final answer is 9. That's a bit of a problem ;)

Suppose n=6. The final answer should be 6 + 4 + 2 = 12.

In your code, if n=6, the final answer is 9. That's a bit of a problem ;)

Try something like this ((n *n)/4 + (n* 0.5))
if n = 6
then ((6*6)/4 + (6 *0.5)) = 12

6*(6/4+0.5) = 6 * 2 = 12

This works for human, this does'nt work in c# , so if you try something like

((n *n)/4 + ( n * 0.5)) = 12 in c#.

Indeed.

This works for human, this does'nt work in c#.

You are correct, because it is integer division!

It would wouk if he wrote n / 4.0.
:D

Well, as it happens 1.5, can be accurately reprepresented in binary - but I do take your point. Rounding errors are always an issue in floating point calculations.
:thumb:

Indeed.

You are correct, because it is integer division!

It would wouk if he wrote n / 4.0.
:D

You are right. It was just a typo mistake of me because I did not test the code. It means n / 4.0 on that part.

But I am understand the point of view for a rounding error. There is a simple solution. You have only to eliminate the fraction inside.

public uint Calculate(uint n)
{
//ensure the number is an even number
if (n % 2 == 1)
n -= 1;
//calculate sum without a loop because it is a arithmetic progression
return n*(n + 2) >> 2;
}

That's it. There should no rounding error any longer. :wave: