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

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The proof of the equation F^2_n + F^2_(n+1) = F_(2n+1) for n ≥ 1 can be established using mathematical induction. The base case is verified by substituting n = 1, confirming that the equation holds true. The inductive step involves assuming the equation is valid for n = k and demonstrating its validity for n = k + 1. This method guarantees the formula's truth for all integers n greater than or equal to 1.

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StellaLuna
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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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