Question

Prove the statement by mathematical induction.

3 + 5 + 7 + . . . + (2n + 1) = n(n + 2)

Answer 1

@terenzreignz Would be use \(1 + 3 + 5 + 7\cdots (2n-1) = n^2\)?

Answer 2

we*

Answer 3

Why ask me? :D
I'm just waiting for you... we might have different approaches :)
(NOTE: Whatever approach you're using, it's not the same as what I did ^.^ )

Answer 4

Besides, instructions clearly say "use induction" so I did just that :D

Answer 5

OK, write down your proof lol

Answer 6

Oh you -.-

Answer 7

Mathematical Induction...
BASE STEP
Show it holds for n = 1...

Well, clearly...
2n + 1, when n = 1
= 3
=n(n+2) when n = 1
=1(1+2) = 3

Answer 8

Thank you both for helping by the way!! :)

Answer 9

Okay...
INDUCTION STEP
You assume it holds for n = k
Then you try to show that it holds for n = k+1

Can you do that, @theequestrian ? :)
And that goes for you too... @ParthKohli -.-

Answer 10

Super quick question. How do you know n=1 & n-k & n=k+1?

Answer 11

Those are the usual steps for proving by Mathematical Induction
Step 1, show that it holds true for n = 1
Step 2, ASSUME it holds true for n = k, and then show it holds for n = k+1

Answer 12

Ohh. Okay. So do you just substitute k for n?

Answer 13

Having said that... we assume, therefore, that it holds for n = k
In other words, we assume that this equation is true...
\[\large 3+5+...(2k+1)=k(k+2)\]

Answer 14

Yes, @theequestrian :)

Answer 15

Oh lol, I got confused with that equation

Answer 16

Humans make mistake. I've become a little too much of a human.

Answer 17

Okay, well thats simple enough so far! Haha so our second step would be..
3 + 5 + 7 + ... + (2k + 1) = k(k + 2)

Answer 18

mistakes*

^ another one -_-

Answer 19

Wouldn't it be proving that\[3 + 5 + 7 + \cdots 2k+1 + 2(k+1) + 1 = (k+1)(k+3)\]

Answer 20

Is that so bad, @ParthKohli ? :P

We assume that this
\[\large 3+5+...(2k+1)=k(k+2)\]is true, and test this out...


\[\large 3+5+...(2k+1)\color{green}{+[2(k+1)+1]}=(k\color{green}{+1)}[(k\color{green}{+1})+2]\]

Answer 21

Understood, @theequestrian ?
@ParthKohli You may have a different way of using PMI... do you assume
it holds for n = k-1 and show that it holds for n = k instead?

It looks to me like you're getting things mixed up...

Answer 22

Yeah, I'm getting them mixed up -_-

Answer 23

I'm actually a little lost..

Answer 24

I quit. :-P

Answer 25

Okay, let's get you up to speed... we assume it is true for n = k, am I right?
So we replace n with k...
\[\large 3+5+...(2k+1)=k(k+2)\]

Answer 26

Okay, I get that. Thats about all I get.. Lol & thank you!

Answer 27

Okay. Next we have to show that it holds for n = k+1...
so you see this?
\[\large 3+5+...(2\color{red}n+1)=\color{red}n(\color{red}n+2)\]

I just replaced all instances of n with a k+1, like so...
\[\large 3+5+...[2\color{red}{(k+1)}+1]=\color{red}{(k+1)}[\color{red}{(k+1)}+2]\]
Now do you understand? :)

Answer 28

Ohhh. So you took the original equation and substituted for k & then went BACK to the original and substituted for (k+1)? What was the point of putting k in alone?

Answer 29

The point, we assumed that the one involving k (alone) is already true, and now, all we have to show (prove) is that the one involving k+1 is ALSO true, given that the one involving k (alone) is true...

get it?

Answer 30

Yes, thank you! & I have to fill in some blanks & I've done some of them, can you check them?

Answer 31

What? Is this the same question?

Answer 32

Yes it is

Answer 33

All right...

Answer 34

I was trying to figure it out & then fill in the blanks, but its actually harder than I had planned

Answer 35

for slot a3, you should know better... remember, it's just the original equation you were trying to prove, save for the fact that we replaced all instances of n with k...

Answer 36

Ohh, so for that line its just simply replacing all the n's with k's?

Answer 37

Oh duh because it says k=n

Answer 38

Yes. -.-

Answer 39

So would
a4 = 2
a5 = k
a6= 2
a7 = 1
I think?

Answer 40

a5? no... look at this more carefully... this is the n=k+1 part, right?
\[\large 3+5+...\boxed{[2\color{red}{(k+1)}+1]}=\color{red}{(k+1)}[\color{red}{(k+1)}+2]\]

Answer 41

It's
[2(k+1) + 1]
right?
Mind simplifying that for me? :P

Answer 42

2k+ 2 +1
2k+3

Answer 43

2k+3 is my answer haha

Answer 44

Then, a5 should instead be...?

Answer 45

it should be 3

Answer 46

Yes :)
Note that a6 and a7 basically mirror a4 and a5

Answer 47

Oh okay, so it would be
a6 = 2
a7 = 3

Answer 48

Yup :)

Answer 49

Thank you soo much!! I kind of get it now:) Couldn't say that before! I really appreciate all your help! Thanks!

Answer 50

No problem :)