Question

Use the Principle of Mathematical Induction to show that the given statement is true for all natural numbers n.

Answer 1

I know n=1 is 3 for both sides.
n=k is 1/2 (k)(k+5)
But I can't get n=k+1

Answer 2

okay. So if you sub in k for k+! what would you end up with?

Answer 3

I'm getting (1/2) (k) (k+5) + (k+1) +2

Answer 4

and on the other side (1/2) (k+1)(k+1)+5

Answer 5

No. What I'm asking you to do is to tell me what the target is. You need a target. The target is when you sub in k for k+1.

Answer 6

@toadytica305 are you there?

Answer 7

All I'm asking you to give me is by giving me your expression when you substitute "k" for "k+1". That's all.

Answer 8

Then we can go from there.

Answer 9

@aztech yes, my internet was acting up.

Answer 10

wouldn't that be (1/2) (k) (k+5)

Answer 11

First off, it's best to work with the LHS to get to the RHS.

Answer 12

So For n=k+1, your target is the RHS with k substituted with k+1.

Answer 13

I have it set up like this, but after this I am stuck..

Answer 14

Okay. I will skip the first two steps or so, and get to n=k.
\[\large 3+4+5+...+(k+2)=\frac{1}{2}k(k+5)\]
Your target is this. YOu want the LEFT hand to become this when n=k+1:
\[\large Target=\frac{1}{2}(k+1)(k+6)\]

Answer 15

correct

Answer 16

Now for n=k+1. The left hand side would become this:
\[\large LHS=3+4+5+...+(k+2)+(k+3)\]

Answer 17

k+3 is the "k+1"th term of the sequence.

Answer 18

k+2 is the "k"-th term of the sequence.

Answer 19

Now, we use the previous expression for n=k.
\[\large 3+4+5+...+(k+2)=\frac{1}{2}k(k+5)\]
So that whole sequence can be substituted with the RHS.

Answer 20

so now I can replace \[3+4+5+.. (k+2) with 1/2 (k) (k+5)\]

Answer 21

\[\large LHS=\frac{1}{2}k(k+5)+(k+3)\]

Answer 22

Are you up to this stage. Or are you confused about anything I'm doing right now?

Answer 23

Okay, I see you're on the right track. Okay let's continue. My computer's a bit slow today so bear with me.

Answer 24

that's what I've gotten up to every time I try it , I don't know what to do to get it to be (1/2)(k+1)(k+6)

Answer 25

Okay the target has 1/2 in it so we take that out as a "common factor"
\[\large LHS=\frac{1}{2}[(k(k+5)+2(k+3)]\]

Answer 26

We don't want to get rid of the 1/2. We need to keep it.

Answer 27

That's the key.

Answer 28

Now we expand the the inside and we get a quadratic.

Answer 29

Are you with me? Or stuck on where the 2 came from beside (k+3)?

Answer 30

that's where I get \[(1/2) (k^2+6k+3)\]

Answer 31

nope, I'm lost on where the 2 came from..

Answer 32

Ah okay.
If you expand that out, where I factorised the 1/2 out. You get your original equation. AM I right?
\[\large \frac{1}{2}[k(k+5)+2(k+3)]=\frac{1}{2}k(k+5)+\frac{1}{2}(2)(k+3)\]

Answer 33

Since your original equation does not contain a contain a 1/2 beside k+3. It was just k+3 originally.

Answer 34

So we need to "cancel out" the half by putting a 2 beside the k+3 when we take 1/2 as a common factor.

Answer 35

Oh okay, got it now

Answer 36

Yep, so now we expand the inside of the brackets.
\[\large LHS=\frac{1}{2}[k(k+5)+2(k+3)]\]
That was our previous line.
\[\large LHS=\frac{1}{2}[k^2+5k+2k+6]\]

Answer 37

and now I get (1/2) (k^2 + 7k +6 which simplifies into (k+1) (k+6)

Answer 38

Yep. Well done. which is the same form. So we've done it!

Answer 39

Good job!

Answer 40

thankkkk you!

Answer 41

Now you can conclude with your last statements.

Answer 42

I never thought about the 1/2 as the common denominator so that killed me every time

Answer 43

and you've completed you mathematical induction of this question. And I'm glad I could help. :)

Answer 44

Here's a secret for you to keep when doing mathemetical inductions. First secret is to write down your target like I did.

Answer 45

That's an important step. Now you look at your target and compare it to the LHS you're dealing with. If you see something similar, you instantly think that you need to take that out as the "common factor" Then you just make certain adjustments to fit that common factor.

Answer 46

that is actually really helpful! I have to solve 4 more mathematical inductions, so that comes in handy.. thank you once again!

Answer 47

No worries.