Math Is Fun Forum

FP1Student

Guest

Induction. o_O

Hey, I'm a little stuck on my induction work. I've managed most of the summing series one, but I get really confused when I get to inequalities.

Okay, I need to prove by mathematical induction that:

What I have got so far is that:

Step 1

Let n=1. LHS is 2x1=2 and the RHS is 1, so it is true as 2>1

Step 2

If n=k, ie. 2+4+6+8+...+2k>k² is true then when n=k+1, it would be:

2+4+6+8+...+2k+2>k²+2k+2

But then how do you go from there?

And there are a few other questions that I am stuck on, but I'm going to try and look over them again.

Thank you in advance.

TheDude

Member

Registered: 2007-10-23

Posts: 361

Re: Induction. o_O

You didn't do the induction step correctly.  It should be:

2 + 4 + 6 + 8 + ... + 2k + (2k + 2) > (k + 1)^2 = k^2 + 2k + 1

You already know that 2 + 4 + 6 + ... + 2k > k^2, so subtract both sides from the inequality.  The proof is trivial from there.

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Wrap it in bacon

mathsmypassion

Member

Registered: 2008-12-01

Posts: 33

Re: Induction. o_O

2+4+....+2k+(2k+2)>K² + (2k +2) = (k+1)² +1 >(k+1)² QED

FP1Student

Guest

Re: Induction. o_O

You didn't do the induction step correctly.  It should be:

2 + 4 + 6 + 8 + ... + 2k + (2k + 2) > (k + 1)^2 = k^2 + 2k + 1

You already know that 2 + 4 + 6 + ... + 2k > k^2, so subtract both sides from the inequality.  The proof is trivial from there.

So eventually from that you end up with 1>0?

TheDude

Member

Registered: 2007-10-23

Posts: 361

Re: Induction. o_O

We know:

2 + 4 + 6 + ... + 2k > k^2

Add 2k + 2 to both sides:

2 + 4 + 6 + ... + 2k + (2k + 2) > k^2 + 2k + 2 > k^2 + 2k + 1 QED.

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Wrap it in bacon

FP1Student

Guest

Re: Induction. o_O

Ah, I'm so silly. >_<

Hmm, this is another inequality one, please tell me if I'm along the right path. Thank you. ^_^

Prove by mathematical induction that

Step 1:

Let n=1, 1^2 = 1 on the LHS and then on the right hand side 1^3/3 = 1/3 which is true as 1>1/3

Step 2:

Assuming n=k is true, then it would become k² > k³/3

Then n=k+1 would be k²+(k+1)²>k³/3+(k+1)³/3
                             ⇒ 6k²+6k+3>2k³+3k²+3k+1

In fact, I'm getting lost. o_O Is that right so far? Do I continue simplifying and cancelling that out?

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Induction. o_O

Assuming n=k is true, then it would become k² > k³/3

No. It would become

Last edited by JaneFairfax (2009-01-28 08:13:43)

FP1Student

Guest

Re: Induction. o_O

Ah, I need to add in the numbers before. Okay then.

Did you mean k³/3 though for the right hand side?

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Induction. o_O

Yes, it’s a cube on the RHS. I’ve corrected it.

FP1Student

Guest

Re: Induction. o_O

I'm really stuck on this. >_< What I've just done is this:

When n=k

And so when n=k+1

And I'm sure that that is not what you're supposed to do.

I'm not even sure if I have grasped this induction concept yet. Please could someone explain?

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Induction. o_O

At times like this it helps to try and work backwards and see if it leads you somewhere. Do this as rough work.

Now you want to get the RHS into the form

which is

So far you only have

So comparing the two, it appears that you want to have

But this is true because

(since

is a natural number) and

and so

.

So now you can continue.