>> Let's use math induction to prove this conjecture. It says that the sum of n items given by 1 plus 2 plus 3 plus 4 and so on through the nth item is equal to the expression n times n plus 1 over 2. Is that true? Well, let's just -- let's see what this means. If we're talking about adding 4 items, 1 plus 2 plus 3 plus 4, that would be 4 and 3, 7 and 2, 9 and 1, 10, so that sum would be 10. Does it meet the form over here if n is 4? Well, this would be 4 times 4 plus 1 over 2, 4 times 5 is 20, 20 over 2, 10, so it certainly works when n is 4. Now, by math induction here's what we need to do. We need to show that the conjecture is true when n is 1. So, when we're adding just one item at the sum of 1 item would be 1. Now, does it meet the form that we have over there? Well, that would be 1 times 1 plus 1 over 2. Well, this is 2 over 2 or 1, so it certainly works in that case. Now, we then assume the truth that the sum of k items meets the pattern that we talk about that we have in the conjecture, but does that imply the truth for the k plus oneth item. That is, is it therefore the truth that the sum of k plus one items given by all of these plus the next item, you see equal to this expression for the k plus oneth item. Now, here's where we have to think about how to write this and where this comes from. This the k plus oneth expression and just replace this k with k plus 1, replace that with k plus 1, you see and we get this, okay, so that's the idea. Now, in order to show this the strategy generally is to write this and show that and then make a replacement, refer back to something that you made in this assumption. Alright, and then we want to lead to this expression, so that's the idea. So, we start with the sum of k plus one terms would be 1 plus and so on all the way to adding the k plus oneth item. Alright, and then we utilize this information and then try to show that, alright, here's what we're doing. We take s of k alright, here's s of k, now s of k plus 1 means just tack on the k plus oneth item, that's what this is. Alright, now here's where we jump start the whole thing. We take this part of it right here and we replace with that, you see that expression. Okay, so that's what comes down to right here and replaces this. And now we still have the k plus oneth item being added. Now, we want to manipulate this expression into the form that we wanted then that form was k plus 1 times k plus 2 over 2, okay, let's see if we can do that. We see that in these two terms k plus 1 is a common factor. So, we pull out k plus 1 as a common factor. When we pull k plus 1 out of this term we're left with k over 2. When we pull k plus 1 out of this term we're left with 1. Now, what about this? What's this? Well, let's see. We have a fraction here we need common denominators, one can be thought of as 2 over 2, so when we write 1 as 2 over 2 we have k plus 2 over the common denominator of 2. So, this becomes that and now we can rearrange factors a little bit and write this as k plus 1 times k plus 2 over 2, which is what we need to do.