# Thread: Help with Proofs by Induction

1. ## Help with Proofs by Induction

Hello everyone. I am really struggling to wrap my head around proofs by induction. I have a problem that I need help understanding.

1. Prove: 4 + 8 + 12 +...+ 4n = 2n2 + 2n for all integers n >= 1

Does the base case always have to equal 1? Or is the base case 4. After doing some algebra I ended up with 4.

After the I get the base case I'm having trouble understanding how to come up with the inductive hypothesis and how to put it all in a proof.

2. ## Re: Help with Proofs by Induction

Originally Posted by troe
After the I get the base case I'm having trouble understanding how to come up with the inductive hypothesis and how to put it all in a proof.
I recommend studying this description of a proof by induction.

3. ## Re: Help with Proofs by Induction

Originally Posted by troe
Hello everyone. I am really struggling to wrap my head around proofs by induction. I have a problem that I need help understanding.

1. Prove: 4 + 8 + 12 +...+ 4n = 2n2 + 2n for all integers n >= 1

Does the base case always have to equal 1? Or is the base case 4. After doing some algebra I ended up with 4.
You ended up with 4 for what? The problem says "for all integers n>= 1". You are doing induction on n.
when n= 1, you have 4= 2(1)2+ 2(1) which is true.

Assume that, for some k, 4+ 8+ 12+ ...+ 4k= 2k2+ 2k. Then the k+1 case is 4+ 8+ 12+ ...+ 4k+ 4(k+1)= 2k+ 2k+ 4k+ 4
Can you show that that is equal 2(k+1)2+ 2(k+1)?

After the I get the base case I'm having trouble understanding how to come up with the inductive hypothesis and how to put it all in a proof.