00:02
In this problem, we are given multiple choices, and we are asked to pick the choice, which is correct.
00:14
So the question that we're asked to answer is, which of these three methods of mathematical proof? i'm going to use to prove the statement of the sum of the first n -odd natural numbers.
00:37
1 plus 3 plus 5 plus dot -dot plus 2 -n minus 1 is equal to, n squared for every natural number n.
00:49
Three methods of proof are one, mathematical induction, two, sum of a finite arithmetic sequence, and three, the sum of a finite geometric sequence.
01:06
So in order to identify which combination of these methods of proof is the correct one, let's go through each method of proof.
01:20
So could we prove this statement? by mathematical induction.
01:26
I would say to start looking at this from an intuitive point of view, this is a perfect statement to prove by mathematical induction because you see that it's a statement about natural numbers, in particular all natural numbers.
01:42
And we've already seen proofs of statements, like some of the first n natural numbers, some of the first and even natural numbers.
01:51
And so we would figure that it should be easy to figure out the sum of the first and odd natural numbers as well.
02:01
However, to ensure that mathematical induction does work to prove this, i wouldn't recommend doing this in multiple choice test, but if you had time, i would work out the proof of this statement.
02:16
So i'm going to do this below.
02:21
So for method one, we have our base case.
02:31
This is the statement p1, which is that, 1 is equal to 1 squared.
02:40
For our guess, instead of 1, we could write 2 times 1 minus 1 equals 1 squared.
02:51
So what is the left side of this equation? we have 2 times 1 minus 1 is going to be 2 minus 1 or 1, and 1 squared is obviously 1.
03:01
So we have the two sides of the equation are the same, and therefore the statement p1 is true.
03:10
For the inductive step, we'll suppose that that, that the statement pn, which is that the sum from 1 to n minus 1 equals n squared, is true for some natural number n.
03:37
We want to show that the statement pn plus 1 is also true then.
03:44
So the left side of the statement pn plus 1 is 1 plus 3 plus 2n minus 1 plus 2n plus 2 times.
03:58
Times n plus 1 minus 1 and by the associativity of addition you can draw brackets around the first n terms of the sum and using fact that our inductive hypothesis is tn this is going to be equal to n squared plus two times n plus 1 minus 1 by the inductive hypothesis which is equal to n squared plus 2n plus 2 minus 1 is plus 1.
04:58
So the question is how do we get this into the form n plus 1 square? recognize that this polynomial is actually just the expansion of n plus 1 square.
05:14
And this is exactly what we wanted to show, prove that the statement tn plus 1 is true.
05:28
And putting this together, we have that if pm is true for some natural number n, then we have that pm plus 1 is also true.
06:13
And therefore, by mathematical induction, it follows that pn is going to be true for all natural numbers.
06:30
And so we have that one is a valid method of proof for this statement...