This is clearly the same thing as k + 2. Equation 17 Finally, we reintegrate the last term into the summation by changing t to t+1 (above the sigma). Well we have already proven that it works for 1 so we can assume it works for 1. ⊕ 4. Since someone decided to revive this 6 year old question, you can also prove this using combinatorics. And so we can try this out with a few things, we can take S of 3, this is going to be equal to 1 plus 2 plus 3, which is equal to 6. Why is this interesting to us? Now what I want to do in this video is prove to you that I can write this as a function of N, that the sum of all positive integers up to and including N is equal to n times n plus one, all of that over 2. Shove the current value of \(n\) into \(k\), evaluate, and do the same for the next value of \(n\), etc. This is the same thing as k plus 1, that's this part right over here. This is what we need to prove. NOTE: \(k\) is a placeholder variable representing each number in a series starting from \(0\), and up to and including \(n\). If the formula to prove is not given in the problem, it can usually discovered by evaluating the rst few cases. But instead of tacking on \(\mathbf{(n+1)^{2}}\) to the RHS (we'll do this in the next step, the proof itself) as we did to the LHS, we're going to use the RHS of Step 2, and carefully insert and replace all instances of \(n\) with \(n+1\), and simplify. If they're equal, then you've got it, and can finish the proof with writing the cubic polynomial in factored form. Right? Sidenotes here and inside the proof will provide commentary, in addition to numbering each step of the proof-building process for easy reference. Our inductive assumption is: Assume there is a k, greater than or equal to zero, such that a k = (1 - 1/2 2 k)/2. We've just added all of them, it is just 1. Right? Proof by mathematical induction. So if you factor out a k + 1, you get k plus 1 times refractoring out over here, if you factor out k + 1 you'd just have a k. Over here if you factor out k + 1 you would just have a 2. Use induction to prove that We are actually showing that the original formula applies to k+1 as well. Use induction to prove that ⊕ Sidenotes here and inside the proof will provide commentary, in addition to numbering each step of the proof-building process for easy reference. This proves the result for \((n+1)\), so the result is true for all \(n \ge 0\) by induction. But since we can assume it works for 2 we can now assume it works for 3. Free Pre-Algebra, Algebra, Trigonometry, Calculus, Geometry, Statistics and Chemistry calculators step-by-step 1. The 2's would cancel out, I'd just wrote it this way so I have a common denominator. \(\Box\) The RHS is simply plug and chug. So we are going to have k times k plus 1 plus 2 times k plus 1. ⊕ Since the formula claims to work for all numbers greater than or equal to (\(\ge\)) \(0\), \(0\) must be tested on both sides. 1. The proof is finished with a concluding statement. And the reason why this works is - Let's say that we prove both of these. The inductive hypothesis is Under this assumption, • Show that the sum of first n positive odd numbers is n2 . This is equal to. we have proven it for 1. And now we can prove that this is the same thing as 1 times 1 plus 1 all of that over 2. Proof Sidenotes here and inside the proof will provide commentary, in addition to numbering each step of the proof-building process for easy reference. If pressed for time on a test and can't factor the cubic polynomial quickly, just go back to Step 3, F.O.I.L. Then in our induction step, we are going to prove that if you assume that this thing is true, for sum of k. If we assume that then it is going to be true for sum of k + 1. Times k plus 1 plus 1. so this is k times k plus 1 over 2 plus 2 times k plus 1 over 2. And we are done. Replace each summation by the closed form given above. Theorem: The sum of the first n powers of two is 2n – 1. They are not part of the proof itself, and must be omitted when written. When it's true for 2, then it must be true for 3, because we have proved it, when it's true for k, it's true for k + 1. To reinsert the first term into the summation formula, we change k=0 to k=1. So this is going to be equal to... We have a common denominator of 2 and I'll write this in a different colour here. Thus, we need to prove an+1 + an+1 −1 a−1 = an+2 −1 a−1. So the base case we're going to prove it for 1. Let me colour code those. Example 1. So you would know what I'm doing. But in this case, we are saying this is true for all positive integers. The left-hand side (LHS) reads, "the sum from k equals zero to n of k squared.". 1 plus 1 is 2, 2 divided by 2 is 1, 1 times 1 is 1. For \(n=0\), the left-hand side (LHS) yields: So the equation holds on both sides for \(n=0\). So we will just take this part and we will add it to k plus 1. so we'll add it to k plus 1 over here. The rest is algebra and simplification. If we assume that this is true and if we use that assumption we get that the sum of all positive integers up to and including k + 1 is equal to k + 1 times k + 1 + 1 over 2. So we showed , we proved our base case. Assume the result for \(n\). We have proven it for 1 and we have proven it that if it works for some integer it will work for the next integer. $$\sum_{k=0}^{n} k^{2} = \dfrac{n(n+1)(2n+1)}{6}$$ for all \(n \ge 0\). Amazing! The hypothesis of Step 1) -- "The statement is true for n = k" -- is called the induction assumption, or the induction hypothesis. ⊕ For the LHS and RHS to stay equal to each other, any changes made to one side must also be made to the other side. That number would be \(\mathbf{(n+1)}\), or the "next thing" we'll try to coax out from the "current thing." Notice in the next to last line of the proof, the result must look the same as what was found in Step 3. Because Sigma's spelled with an "s", it's used to represent a summation, or sum. Notice in the next to last line of the proof, the result must look the same as what was found in Step 3. Now what I want to do is think about what happens when I try to find this function for k + 1. The rest is algebra and simplification. If they're equal, then you've got it, and can finish the proof with writing the cubic polynomial in factored form. So if it's true for 2 it's true for 3. and if it's true for 3 it's going to be true for 4. Proof by Induction for the Sum of Squares Formula. So what is the sum of all of the integers up to and including k + 1? 1, 2, 3, 4, 5, 6 you could just keep going on forever. Now spoken in generalaties let's actually prove this by induction. 1 Induction The idea of an inductive proof is as follows: Suppose you want to show that something is true for all positive integers n. (The catch: you have to already know what you want to prove — induction can prove a formula is true, but it won’t produce a formula you haven’t already guessed at.)