Need help with verifying if a proof i did is correct.

Question

i am tasked with proving the following : Given that $$a_{1}=1,a_{2}=2$$ and $$a_{n}=a_{n-1}+a_{n-2}\;(1)for \;n\ge 3 $$ prove that : $$a^{2}_{n}-a_{n-1}a_{n+1}=(-1)^{n}\; for \; every \; n \ge2$$ For n=2 this is true because both sides give 1 .For n >2, i tried proving it by induction by assuming that in the equality that i want to prove , n=k holds , so i tried showing that n=k+1 holds . So far i have done the following : $$for \;n=k : a^{k}_{2}-a_{k-1}a_{k_+1}=(-1)^{k}(assumption)(2)$$ $$for \; n=k+1: $$ $$a^{2}_{k+1}-a_{k}a_{k+2} = (-1)^{k+1}$$ $$-a^{2}_{k+1}+a_{k}a_{k+2} = (-1)^{k} (3)$$ If i put n=k+2 in (1) i get the following : $$a_{k+2}=a_{k+1}+a_{k}(4)$$ So i plug (4) in (3): $$-a^{2}_{k+1}+a_{k}(a_{k+1}+a_{k}) = (-1)^{k}$$ $$-a^{2}_{k+1}+a_{k}a_{k+1}+a_{k}^{2} = (-1)^{k}(5)$$ Then i plug (2) into (5): $$-a^{2}_{k+1}+a_{k}a_{k+1}+a_{k}^{2} = a^{k}_{2}-a_{k-1}a_{k_+1}$$ $$-a^{2}_{k+1}+a_{k}a_{k_+1}+a_{k-1}a_{k+1}=0$$ Then i end up here: $$a_{k+1}(-a_{k+1}+a_{k}+a_{k-1})=0(6)$$ Which means that what is inside the parenthesis must be equal to zero, which is true , since in (1) if we say n = k + 1 we will end up with this : $$a_{k+1}=a_{k}+a_{k-1}$$ which is the same thing with what's inside the parenthesis in (6) if we solve for 0.

Is this correct ? If not , can you tell me what i did wrong and how you would go about proving this?

Thank you very much!

Answer 1

The proof looks good. There are just a few places where you have a typo. Maybe you should try rewriting in a way its less complicated to read. In your proof are many references like (i). But that is complaining on high level. Good job

Furtheremore you shouldn't write n=k holds and we want to show for n=k+1. Just: We assume our Satement holds for n and want to prove the Statement for n+1. That's enough and you could avoid this extra variable k.