Download Problem 4: Show that: 1 + 2 + 3 + + n = (n2+n) /2.

Problem 4:
Show that: 1 + 2 + 3 + ... + n = (n2+n) /2.
Prove the Base Case:
The sum of 1 is 1. This we establish through our own reasoning.
The equation tells us that: (12+1) /2 = (1+1) /2 = 2/2 = 1
Thus, the equation works for the base case.
Make the induction hypothesis
Assume for arbitrary n that the sum of numbers up to the nth number is (n2+n) /2
Prove the induction step
The next step after n is n+1.
Since we assume in the induction hypothesis that the sum of numbers up to nth number is
(n2+n) /2, then the sum up to the n+1th number is: ((n2+n)/2) + (n+1).
So our task is to show that this is equivalent to the proposed theorem, for the n+1th case.
So, substitute in (n+1) for n in the equation (n2+n) /2. That gives us: ((n+1)2+ n+1) /2.
So, show that ((n2+n)/2) + (n+1) = ((n+1)2+ n+1) /2 and we will have proven the
induction step.
Consider the left side. We will reason through our observation (we are not using the
equation, except in the induction hypothesis). We see:
((n2+n)/2) + (n+1) = ((n2+n)/2) + (2n+2)/2) = ((n2+n) + (2n+2))/2 = (n2+3n+2)/2
Consider now the right side. Here, we are applying the equation. We are hoping it will
come out the same as our independent reasoning above. We see:
((n+1)2+ n+1) /2 = ((n2 + 2n + 1) + n + 1)/2 = (n2 + 3n + 2) /2
So, since the two are identical, we have proven the equation applies in the n+1th case, and
so we have proven the induction step.
Conclusion
Using mathematical induction we now conclude that the sum of numbers up to the nth
number, for any n, is always (n2+n) /2.