Prove by Induction that 4 + 8 + 12 + ... + 4n = 2n^2+2n

[ezwind72]

I'm not sure if this is the correct section for this problem, if not, I'm sorry. I'm not even sure anybody can help me with this. I need to prove by induction that 4+8+12+...+4n=2n^2+2n for all integers n is greater than or equal to 1. We did a problem like this in class 2 weeks ago and I had a very difficult time understanding his work. He said it was simple if we just thought about it and that it was basically just 2 steps. Yet it took him 10 minutes to prove it. I asked several questions but I honestly do not have a clue what he is talking about. Is there anyone out there that can tell me how to clearly do this??

1) He said to first substitute a number in for n, I used 1.
2) Then he said to make an assumption and change the n to a k so I get 4+8+12+...4k=2k^2+2k.

This is where it gets hairy.
3) now I think I am supposed to do 4+8+12+...4k+(4k+1)= I'm lost!

 

[Deleted member 4993]

Re: Prove by Induction

2) Then he said to make an assumption and change the n to a k so I get 4+8+12+...4k = 2k^2+2k = 2k(k+1).

This is where it gets hairy.
3) now I think I am supposed to do 4+8+12+...4k + 4(k+1) = 2k^2+2k + 4(k +1) = 2(k^2 + 3k +1) = 2(k+1)[(k+1)+1]

and continue

 

[ezwind72]

Re: Prove by Induction

2) Then he said to make an assumption and change the n to a k so I get 4+8+12+...4k = 2k^2+2k = 2k(k+1).

This is where it gets hairy.
3) now I think I am supposed to do 4+8+12+...4k + 4(k+1) = 2k^2+2k + 4(k +1) = 2(k^2 + 3k +1) = 2(k+1)[(k+1)+1]

and continue

Ok, I think I see what you did there and that is a tremendous help. But am I wrong, or is it supposed to be 2(k^2+3K+2)? So when all is said and done, I would take 2(k+1)[(k+1)+1] and I would multiply that together to get 2(k^2+3k+2)=2(k^2+3k+2)? And that proves it, correct?

 

[Deleted member 4993]

Re: Prove by Induction

... is it supposed to be 2(k^2+3K+2) <<< Correct

So when all is said and done, I would take 2(k+1)[(k+1)+1]

This is the same form - that was given to you for k'th term - now it is for (k+1)th term.

 

[ezwind72]

Re: Prove by Induction

So when all is said and done, I would take 2(k+1)[(k+1)+1]

This is the same form - that was given to you for k'th term - now it is for (k+1)th term.

and I would multiply that together to get 2(k^2+3k+2)=2(k^2+3k+2)? And that proves it, correct?

OK, I am confused again. How come we made 2(k^2+3k+2)=2(k+1)[(k+1)+1] and not 2(k+1)(K+2)? And when all is said and done is this supposed to equal 2k(k+1)? I'm sorry, I must be really frustrating you, b/c I am really frustrating myself. I think maybe I am making it harder than it needs to be.

 

[galactus]

Re: Prove by Induction

If we add 4(k+1) to both sides, we get:

\(\displaystyle 2k^{2}+2k+4(k+1)\) on the right side.

Which, upon a little algebraic manipulation, gives us \(\displaystyle 2(k+1)^{2}+2(k+1)\)

Which is what we want to show.

 

[ezwind72]

Re: Prove by Induction

If we add 4(k+1) to both sides, we get:

\(\displaystyle 2k^{2}+2k+4(k+1)\) on the right side.

Which, upon a little algebraic manipulation, gives us \(\displaystyle 2(k+1)^{2}+2(k+1)\)

Which is what we want to show.

Oh, I see, that makes sense to me. I thought I was supposed to substitute in 4(k+1) somewhere, but really I'm just adding it to both sides. Wow, thank you both for all of your help! Hopefully when I see this type of problem again on a test, I can prove it correctly. I feel much more confident now.

 

[PAULK]

I'm not sure if this is the correct section for this problem, if not, I'm sorry. I'm not even sure anybody can help me with this. I need to prove by induction that 4+8+12+...+4n=2n^2+2n for all integers n is greater than or equal to 1. We did a problem like this in class 2 weeks ago and I had a very difficult time understanding his work. He said it was simple if we just thought about it and that it was basically just 2 steps. Yet it took him 10 minutes to prove it. I asked several questions but I honestly do not have a clue what he is talking about. Is there anyone out there that can tell me how to clearly do this??

1) He said to first substitute a number in for n, I used 1.
2) Then he said to make an assumption and change the n to a k so I get 4+8+12+...4k=2k^2+2k.

This is where it gets hairy.
3) now I think I am supposed to do 4+8+12+...4k+(4k+1)= I'm lost!

......................................................
Here is the scheme for doing M.I. (BTW, don't ever use those letters while talking to your cardiologist.)

Phase 1. WRITE the theorem for n = the 'base case'; usually, but not always, n = 1.
Check that it is correct.

Phase 2a. WRITE the theorem for n = k. This is the ASSUMPTION.

Phase 2b.WRITE the theorem for n = (k+1). This is the STATEMENT TO PROVE.
.............................
NOW a couple of tricks for handling a proof involving a SUMMATION, like yours:

The SUM for n=1 will almost certainly contain only one term, with no dots. ...

The SUM for n=k+1 will almost certainly CONTAIN the sum for n = k, and just have one extra term. You will replace the sum for n=k with the 'right side' of your ASSUMPTION.

Your theorem is: 4 + 8 + 12 + ... + 4n = 2n^2 + 2n

n = 1: 4 = 2(1)^2 + 2(1) = 2(1) + 2 = 2 + 2 = 4, check.

n = k: 4 + 8 + 12 + ... + 4k = 2k^2 + 2k --- assumption

n = k+1: 4 + 8 + 12 + ... + 4k + 4(k+1) = 2(k+1)^2 + 2(k+1) --- to prove

Note the presence of the 'first k' terms on the left. Replace with the RHS of your assumption, do the algebra, and you have it.

 

[ezwind72]

I get it!! It is all very clear to me now and I now see where I had gone so very wrong. I have come up with 2(k+1)^2+2(k+1) for both sides of my equation. Thank you to everyone, it is a relief knowing that I can do this now!