I recommend studying this description of a proof 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 = 2n^{2 }+ 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.
I recommend studying this description of a proof by induction.
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= 2k^{2}+ 2k. Then the k+1 case is 4+ 8+ 12+ ...+ 4k+ 4(k+1)= 2^{k}+ 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.