Question

Prove the following proposition using the principle of mathematical induction

Answer 1

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

Answer 2

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

Answer 3

yes

Answer 4

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.

Answer 5

We call this the Induction Hypothesis ^

Answer 6

yeah

Answer 7

okay

Answer 8

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

Answer 9

to equal n^2?

Answer 10

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.

Answer 11

is it not k+2?

Answer 12

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

Answer 13

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

Answer 14

k+2? what? :o

Answer 15

i dont get the part (2(k+1)−1)

Answer 16

i dont get where k+1 came from

Answer 17

wait i get it

Answer 18

i get it

Answer 19

XD

Answer 20

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.

Answer 21

yup

Answer 22

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

Answer 23

1+3+...+(2k−1)=k^2

Answer 24

this?

Answer 25

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

Answer 26

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

Answer 27

yeah

Answer 28

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

Answer 29

ohhhhh oh kay

Answer 30

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.

Answer 31

thank yyou!

Answer 32

np \c:/