I'm assuming this is using strong induction/ regular induction. However, besides the "base case" I'm really confused with the inductive steps in my notes. The inductive steps in my notes use the alpha (golden ratio) but I'm confused on how that connects with the proof and how the it's used.
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I wonder why an edit was approved that had $f_n+1$ where $f_{n+1}$ belongs? $\qquad$ – Michael Hardy Mar 25 '16 at 15:41
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Since $F_k=F_{k+1}-F_{k-1}$ for all $k\geq 1$, we have that $$\sum_{k=1}^n F_k^2=\sum_{k=1}^n F_k(F_{k+1}-F_{k-1})$$ $$\implies\sum_{k=1}^n F_k^2=\sum_{k=1}^n F_kF_{k+1}-\sum_{k=1}^n F_kF_{k-1}$$ But since $F_0=0$, we have $S=\sum_{k=1}^n F_kF_{k-1}=\sum_{k=1}^{n-1} F_{k+1}F_{k}$. Then $\sum_{k=1}^n F_kF_{k+1}=S+F_nF_{n+1}$ and finally we get, $$\sum_{k=1}^n F_k^2=F_nF_{n+1}$$ – Sahan Manodya Sep 03 '21 at 13:45
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HINT:
If $\sum_{r=1}^m f^2_r=f_mf_{m+1}+1$,
$$\sum_{r=1}^{m+1} f^2_r=\sum_{r=1}^m f^2_r+f^2_{m+1}=f_mf_{m+1}+1+f^2_{m+1}=1+f_{m+1}(f_m+f_{m+1})$$
$$f_m+f_{m+1}=\text{?}$$
lab bhattacharjee
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@DietrichBurde, Thanks for pointing out. Was that inductive? – lab bhattacharjee Mar 25 '16 at 15:42