Really!
With pleasure!
Here's a simple one.
Hypothesis:
1 + 2 + 3 + 4 + ... + n = n(n+1)/2; for n ∈ Z
+First of all, prove it exists for one case
Case 1: Let n = 1
1 = 1(2)/2
1 = 1
Therefore holds true for n = 1.
Step 2:
ASSUME it works for the kth test where k ∈ Z
+Let n = k
1 + 2 + 3 + ... + k = k(k + 1)/2
Step 3:
Now all we need to do is show it works for the kth + 1 case, seeing as case 1 is true, if we prove it for case k + 1, we can say that if case 1 is true, then case 2 is true. If case 2 is true, then case 3 is true, etc to infinity.
Therefore, let n = k + 1
1 + 2 + 3 + ... + k + (k + 1) = (k + 1)(k + 2) / 2
From step 2 we can reduce LHS with the RHS result in step 2
(k(k + 1) / 2) + (k + 1) = (k + 1)(k + 2)/2
(k(k+1) + 2(k + 1)) /2 = (k + 1)(k + 2) / 2 // Make a common denominator
(k
2 + 3k + 2) / 2 = (k + 1)(k + 2) / 2 // Expand
(k + 1)(k + 2) / 2 = (k + 1)(k + 2) / 2 // factorize
Therefore LHS = RHS and true
Step 4:
As we concluded that as long as the kth test is true, then the (k + 1)th test will hold true. Seeing as the test held through when k was equal to 1, then by induction, all possible tests for k being a positive integer hold true. Proven by mathematical induction!