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anonymous
 one year ago
Prove the following proposition using the principle of mathematical induction
anonymous
 one year ago
Prove the following proposition using the principle of mathematical induction

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zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2\[\large\rm \sum_{i=1}^n (2i1)=n^2,\qquad\forall n\in\mathbb Z^+\] Ok so we start with our base case: \(\large\rm n=1\)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2\[\large\rm (2\cdot 11)=21=1\]Which is equal to \(\large\rm 1^2\), so our base case is satisfied, ya?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2We'll assume it's true for \(\large\rm n=k\). Which means we assume \(\large\rm \color{orangered}{1+3+...+(2k1)=k^2}\) is true.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2We call this the Induction Hypothesis ^

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Then for our Induction Step: \(\large\rm n=k+1\) Let's see if we can get our formula set up correctly :d

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Well, notice that our last number in the sum will be \(\large\rm k+1\) So we won't end up with \(\large\rm n^2\), we'll end up with \(\large\rm (k+1)^2\) on the right. But setting up the left side is going to be a little tricky. Hopefully you can follow what I'm doing here.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Your summation is giving you a bunch of terms like this:\[\large\rm (2\cdot\color{royalblue}{1}1)+(2\cdot\color{royalblue}{2}1)+...+(2\color{royalblue}{k}1)+(2\color{royalblue}{(k+1)}1)=(k+1)^2\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2I'll simplify the first few terms like I did the last time.\[\large\rm 1+3+...+(2k1)+(2(k+1)1)=(k+1)^2\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i dont get the part (2(k+1)−1)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i dont get where k+1 came from

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2So in the Induction Hypothesis, we let \(\large\rm k\) be the largest number in the sequence. We replaced n by k. In our Induction Step, we're letting \(\large\rm k+1\) be the largest number in the sequence. It's the number that comes after k. So our summation grows a little bit, by one more term. And it should be equivalent to (k+1)^2 now, instead of k^2.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2If you scroll up, you'll notice I colored our Induction Hypothesis in orange. We need to relate this back to that orange thing.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Yes, that entire left side is contained within our Induction Step formula.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2\[\large\rm \color{orangered}{1+3+...+(2k1)}+(2(k+1)1)=(k+1)^2\]Do you see it? :)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Since our Induction Hypothesis tells us that the orange stuff is k^2, we can make the switch,\[\large\rm \color{orangered}{k^2}+(2(k+1)1)=(k+1)^2\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2From there, you need to turn the left side into (k+1)^2 somehow. Do some Algebra steps expand out the stuff, then try to factor it down.
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