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

1. anonymous

2. zepdrix

$\large\rm \sum_{i=1}^n (2i-1)=n^2,\qquad\forall n\in\mathbb Z^+$ Ok so we start with our base case: $$\large\rm n=1$$

3. zepdrix

$\large\rm (2\cdot 1-1)=2-1=1$Which is equal to $$\large\rm 1^2$$, so our base case is satisfied, ya?

4. anonymous

yes

5. zepdrix

We'll assume it's true for $$\large\rm n=k$$. Which means we assume $$\large\rm \color{orangered}{1+3+...+(2k-1)=k^2}$$ is true.

6. zepdrix

We call this the Induction Hypothesis ^

7. anonymous

yeah

8. anonymous

okay

9. zepdrix

Then for our Induction Step: $$\large\rm n=k+1$$ Let's see if we can get our formula set up correctly :d

10. anonymous

to equal n^2?

11. zepdrix

Well, 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.

12. anonymous

is it not k+2?

13. zepdrix

Your summation is giving you a bunch of terms like this:$\large\rm (2\cdot\color{royalblue}{1}-1)+(2\cdot\color{royalblue}{2}-1)+...+(2\color{royalblue}{k}-1)+(2\color{royalblue}{(k+1)}-1)=(k+1)^2$

14. zepdrix

I'll simplify the first few terms like I did the last time.$\large\rm 1+3+...+(2k-1)+(2(k+1)-1)=(k+1)^2$

15. zepdrix

k+2? what? :o

16. anonymous

i dont get the part (2(k+1)−1)

17. anonymous

i dont get where k+1 came from

18. anonymous

wait i get it

19. anonymous

i get it

20. zepdrix

XD

21. zepdrix

So 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.

22. anonymous

yup

23. zepdrix

If you scroll up, you'll notice I colored our Induction Hypothesis in orange. We need to relate this back to that orange thing.

24. anonymous

1+3+...+(2k−1)=k^2

25. anonymous

this?

26. zepdrix

Yes, that entire left side is contained within our Induction Step formula.

27. zepdrix

$\large\rm \color{orangered}{1+3+...+(2k-1)}+(2(k+1)-1)=(k+1)^2$Do you see it? :)

28. anonymous

yeah

29. zepdrix

Since our Induction Hypothesis tells us that the orange stuff is k^2, we can make the switch,$\large\rm \color{orangered}{k^2}+(2(k+1)-1)=(k+1)^2$

30. anonymous

ohhhhh oh kay

31. zepdrix

From 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.

32. anonymous

thank yyou!

33. zepdrix

np \c:/

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