proof by mathematical induction

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.

So this formula right over here, this expression it worked for 1, so we have proved our base case. The symbol P denotes a sum over its argument for each natural We then follow that assumption to its logical conclusion. Theorem. 1, 2, 3, 4, 5, 6 you could just keep going on forever. That’s because Series A has an extra a in each term. Initial Step. Now at this step right over here you can factor out a k plus 1. If you would take k + 1 and put it in for n you got exactly the result that we got over here. Based on the inductive hypothesis and Equation 4: We want to add Series A and Series B (Equation 6), but we have a problem. Mathematical Induction is a mathematical technique which is used to prove a statement, a formula or a theorem is true for every natural number.. We’ll apply the technique to the Binomial Theorem show how it works. Since you have verified P(a), it follows from the inductive step that P(a+1) is true, and hence, P(a+2) is true, and hence P(a+3) is true, and so on. And then we are going to prove that if we know it is true for any given k that is true for the next one So if we know it is true for 1 in our base case then the second step, this induction step must be true for 2 then.

We could use n=0 as our base step. And the way I'm going to prove it to you is by induction. https://www.khanacademy.org/.../alg-induction/v/proof-by-induction So we are done with the initial step.

We assume that we have some integer t, for which the theorem works. Definition. We must prove, for this same k, the formula 1 + 2 + ... + k + (k+1) = (k+1)(k+2)/2. Well we have just proven it. We’ll apply the technique to the Binomial Theorem show how it works.

Rather than invoke the Rule, we will derive it for this particular case. We are actually showing that the original formula applies to k+1 as well.

Next, we invoke the inductive assumption, for this k, to get. Because we proved generally if it's true for k it's going to be true for k + 1. Well we have already proven that it works for 1 so we can assume it works for 1. We leave it to the reader to confirm the trivial case of t=0 to complete the proof. The Binomial Theorem tells us how to expand a binomial raised to some non-negative integer power. And now we can prove that this is the same thing as 1 times 1 plus 1 all of that over 2. We combine Equations 12 and 13 to produce a single binomial coefficient (Equation 14). Here we have a sum of t terms with additional terms tacked to the front and back. The proof will now proceed in two steps: the initial step and the inductive step. We are assuming that we already have a formula for this. 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. ak+1 = 2 (1 - 1/22k)/2 (1 - (1 - 1/22k)/2) = (1 - 1/22k)(1 + 1/22k)/2 = (1 - 1/22k+1)/2. Write out the first term, then start the index at k=1 instead of k=0. However, we we extract the first term from Series A and the last term from Series B. The proof involves two steps: But since we can assume it works for 2 we can now assume it works for 3. We started by assuming that the Binomial Theorem applies to some number, t. We have now shown that it follows from that assumption that the Theorem must then apply to t+1.

We are making a general statement about all integers. The elements are recombined to produce a new binomial coeffient, multiplied by the leftover elements.

You see how this induction step is kinda like a domino, it cascades and we can go on and on forever so it works for all positive integers.

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