If n is a positive integer then show that 1+2+3+....+n=n(n+1)over 2
We can prove this formula by mathematical induction.
Base case:
When n=1,
1 = 1(1+1)/2
1 = 1(2)/2
1 = 2/2
1 = 1
This is true for n=1.
Inductive step:
Assume that the formula is true for n=k, i.e. 1+2+3+...+k = k(k+1)/2.
Now we need to show that it's also true for n=k+1, i.e. 1+2+3+...+k+(k+1) = (k+1)(k+2)/2.
Starting with the left side of the equation:
1+2+3+...+k+(k+1) = k(k+1)/2 + (k+1)
= (k(k+1) + 2(k+1))/2
= (k^2 + k + 2k + 2)/2
= (k^2 + 3k + 2)/2
= (k+1)(k+2)/2
Therefore, if the formula holds true for n=k, it holds true for n=k+1.
Since the formula holds true for the base case and for the inductive step, it is true for all positive integers n by mathematical induction.