To prove 1+2+3+...+n=n(n+1)/2
Let S = 1+2+3+...+n
In this case, S is also:
n+n-1+n-2+...+1
If you add both equations of S, you get:
2S=n+1+(n-1+2)+(n-2+3) + ... + n+1
2S = n+1 + n+1 + n+1 + ... + n+1 (n times)
Therefore S = n(n+1)/2
Therefore 1+2+3+...+n = n(n+1)/2
Now for the other two theorems stated in my previous post, how to derive them remains a puzzle to me at this stage. I can prove them via mathematical induction only.
Let S = 1+2+3+...+n
In this case, S is also:
n+n-1+n-2+...+1
If you add both equations of S, you get:
2S=n+1+(n-1+2)+(n-2+3) + ... + n+1
2S = n+1 + n+1 + n+1 + ... + n+1 (n times)
Therefore S = n(n+1)/2
Therefore 1+2+3+...+n = n(n+1)/2
Now for the other two theorems stated in my previous post, how to derive them remains a puzzle to me at this stage. I can prove them via mathematical induction only.
