Prove by induction the sum of n consecutive positive integers is of the form n(n+1)/2.

As proof by induction goes we always have to show that it works for the base case, which in this case is the very first positive integer:1. So we show that the sum of the first number(s) is equal to one which it is as 1=1, then replace n with 1 and we can see that 1(1+1)/2 = 2/2 = 1. So this result is true for n=1, now we use the inductive step and we assume that the result is true for some positive integer k. So the sum 1+2+3+...+(k-1)+k = k(k+1)/2.Now we need to consider the sum of k+1 positive integers (we're looking to find that the sum from 1 to k+1 is equal to (k+1)((k+1)+1)/2. So consider the sum 1+2+3+...+k+(k+1). From our assumption we know that sum from 1 to k is k(k+1)/2 so replace it with that and we get: k(k+1)/2 + k+1, with some rearranging we can find that (1/2)(k^2+k+2k+1) which we can factorise to (k+1)(k+2)/2 which is of the form we want and therefore the proof is complete. Now make a conclusion about it, so if it's true for any positive integer, k , then we've shown for it to be true every k+1, and since it's true for n=1, then it must be true for every positive integer greater than or equal to 1.