Proof of F^2_n + F^2_(n+1) = F_(2n+1) for n>=1

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The discussion centers on proving the equation F^2_n + F^2_(n+1) = F_(2n+1) for n ≥ 1 using mathematical induction. The proof begins with a base case, specifically n = 1, to verify the formula's validity. Following this, an inductive step is proposed where the assumption that the formula holds for n = k is used to demonstrate its truth for n = k + 1. This method establishes that if the formula is true for one case, it holds for all subsequent integers. The conclusion emphasizes that the inductive proof effectively confirms the formula's validity for all n.
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Could anyone help me with the following proof?

F^2_n + F^2_(n+1) = F_(2n+1) for ngreater than or equal to 1?
 
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I think a good technique would be to use induction (I don't know if it actually works, I haven't done the proof).

You first consider a base case, in this case it would be n = 1. Check to see that for this the formula works.

Then comes the inductive step. Assume that this formula works for n =k, and then prove that it works for n = k+1.

Then you're done. The reason this proof works is that truth for n=1 implies truth for n=2, and then n=3, and so on infinitely, so the formula would work for all n.
 

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