Confusion with Algorithm for n + (n-1) + (n-2)

I am trying to expand this answer in iteration after using recursion but i am getting the wrong answer for certain variables.

the question was to write a code for n + (n-1) + (n-2)... till it gets to 3 + 2 + 1

i did it with recursion and it works great but the iteration is the one giving me problem.

this is what i tried in java.

for (int i =1; i <num; i ++) {
  s += n -1;
  }
  return s;

where n is the user inputed varibale, and s is some kind of sum variable.
Also please can you explain what the s+=n -1 means?

i am finding it hard to place it right now. Please and Thank you!

Also i tried to put it <= num
and return s-1; but it did not work either

To do exactly as you described, you can do it like this.

int sum = 0;
int n = 100;
for (int i = 0; i < n; i++) {
   sum += n-i; // first n, then n-1, then n-2, ...
}

But that is the same as the following, only in reversed order.

for (int i = 1; i <= n; i++) {
   sum += i;
}

and as stated in the comments, the arithmetic series sum where the difference between adjacent values is 1, for any value n, is

sum = (n*(n+1))/2;

All the above yield the same result.

looking at your code, you are not even adding the right thing to the s

• you add n-1 but that doesn't even contain your iteration variable i and also, you are iterating to num, not 'n' - you should be adding num-i
• also, check what your s initially is, if you got int s = num; somewhere above, it's ok. But, if you got int s = 0;, you will have to either change it to num or start at i = 0 instead

Let's do it exactly as it's stated: we should loop over n, n - 1, n - 2, ..., 3, 2, 1, so we

• start looping from n: int i = n
• loop up and including 1: i >= 1
• decrement loop valitable (let it be i) by 1: --i

We are ready to write for loop caption:

for (int i = n; i >= 1; --i)

And the code can be

int n = 100;

int s = 0;

for (int i = n; i >= 1; --i)
    s = sum + i; 

If we want to wrap it into method:

static int MySum(int n) {
  int s = 0;

  for (int i = n; i >= 1; --i)
    s = sum + i; 

  return s;
}