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Question Number 208217 by mr W last updated on 07/Jun/24
1^2 +2^2 +3^2 +5^2 +8^2 +13^2 +21^2 =?
$$\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} +\mathrm{8}^{\mathrm{2}} +\mathrm{13}^{\mathrm{2}} +\mathrm{21}^{\mathrm{2}} =? \\ $$
Answered by A5T last updated on 07/Jun/24
This is trivial,but could be interesting with a connection with Fibonacci sequence. f_n =f_(n−1) +f_(n−2) ; f_1 =f_2 =1,f_3 =2 f_1 ^2 +f_2 ^2 +...+f_n ^2 =^? f_n f_(n+1) ...(i) Induction;base case: f_1 ^2 =f_1 ×f_2 =1;f_1 ^2 +f_2 ^2 =f_2 ×f_3 =2 So,suppose ∃n_1 (n_1 =1) such that (i) is true;we show that it must be true for n_1 +1; hence for all n≥n_1 f_1 ^2 +f_2 ^2 +...+f_n_1 ^2 =f_n_1 f_(n_1 +1) ⇒f_1 ^2 +f_2 ^2 +...f_n_1 ^2 +f_(n_1 +1) ^2 =f_n_1 f_(n_1 +1) +f_(n_1 +1) ^2 =f_(n_1 +1) (f_n_1 +f_(n_1 +1) )=f_(n_1 +1) f_(n_1 +2) ⇒n_1 +1 is true; so (i) is true for all n≥1 ⇒1^2 +1^2 +2^2 +3^2 +5^2 +8^2 +13^2 +21^2 =21×34 ⇒?=(21×34)−1=713
$${This}\:{is}\:{trivial},{but}\:{could}\:{be}\:{interesting} \\ $$$${with}\:{a}\:{connection}\:{with}\:{Fibonacci}\:{sequence}. \\ $$$${f}_{{n}} ={f}_{{n}−\mathrm{1}} +{f}_{{n}−\mathrm{2}} ;\:{f}_{\mathrm{1}} ={f}_{\mathrm{2}} =\mathrm{1},{f}_{\mathrm{3}} =\mathrm{2} \\ $$$${f}_{\mathrm{1}} ^{\mathrm{2}} +{f}_{\mathrm{2}} ^{\mathrm{2}} +…+{f}_{{n}} ^{\mathrm{2}} \overset{?} {=}{f}_{{n}} {f}_{{n}+\mathrm{1}} \:…\left({i}\right) \\ $$$${Induction};{base}\:{case}: \\ $$$${f}_{\mathrm{1}} ^{\mathrm{2}} ={f}_{\mathrm{1}} ×{f}_{\mathrm{2}} =\mathrm{1};{f}_{\mathrm{1}} ^{\mathrm{2}} +{f}_{\mathrm{2}} ^{\mathrm{2}} ={f}_{\mathrm{2}} ×{f}_{\mathrm{3}} =\mathrm{2} \\ $$$${So},{suppose}\:\exists{n}_{\mathrm{1}} \left({n}_{\mathrm{1}} =\mathrm{1}\right)\:{such}\:{that}\:\left({i}\right)\:{is}\:{true};{we}\: \\ $$$${show}\:{that}\:{it}\:{must}\:{be}\:{true}\:{for}\:{n}_{\mathrm{1}} +\mathrm{1};\:{hence}\:{for}\:{all} \\ $$$${n}\geqslant{n}_{\mathrm{1}} \\ $$$${f}_{\mathrm{1}} ^{\mathrm{2}} +{f}_{\mathrm{2}} ^{\mathrm{2}} +…+{f}_{{n}_{\mathrm{1}} } ^{\mathrm{2}} ={f}_{{n}_{\mathrm{1}} } {f}_{{n}_{\mathrm{1}} +\mathrm{1}} \\ $$$$\Rightarrow{f}_{\mathrm{1}} ^{\mathrm{2}} +{f}_{\mathrm{2}} ^{\mathrm{2}} +…{f}_{{n}_{\mathrm{1}} } ^{\mathrm{2}} +{f}_{{n}_{\mathrm{1}} +\mathrm{1}} ^{\mathrm{2}} ={f}_{{n}_{\mathrm{1}} } {f}_{{n}_{\mathrm{1}} +\mathrm{1}} +{f}_{{n}_{\mathrm{1}} +\mathrm{1}} ^{\mathrm{2}} \\ $$$$={f}_{{n}_{\mathrm{1}} +\mathrm{1}} \left({f}_{{n}_{\mathrm{1}} } +{f}_{{n}_{\mathrm{1}} +\mathrm{1}} \right)={f}_{{n}_{\mathrm{1}} +\mathrm{1}} {f}_{{n}_{\mathrm{1}} +\mathrm{2}} \\ $$$$\Rightarrow{n}_{\mathrm{1}} +\mathrm{1}\:{is}\:{true};\:{so}\:\left({i}\right)\:{is}\:{true}\:{for}\:{all}\:{n}\geqslant\mathrm{1} \\ $$$$\Rightarrow\mathrm{1}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} +\mathrm{8}^{\mathrm{2}} +\mathrm{13}^{\mathrm{2}} +\mathrm{21}^{\mathrm{2}} =\mathrm{21}×\mathrm{34} \\ $$$$\Rightarrow?=\left(\mathrm{21}×\mathrm{34}\right)−\mathrm{1}=\mathrm{713} \\ $$
Commented by mr W last updated on 07/Jun/24
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Answered by zamin2001 last updated on 07/Jun/24
1^2 +1^2 +2^2 +3^2 +5^2 +8^2 +13^2 +21^2 =21×34 ⇒ 1^2 +2^2 +3^2 +5^2 +8^2 +13^2 +21^2 =21×34−1=713
$$\mathrm{1}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} +\mathrm{8}^{\mathrm{2}} +\mathrm{13}^{\mathrm{2}} +\mathrm{21}^{\mathrm{2}} =\mathrm{21}×\mathrm{34} \\ $$$$\Rightarrow\:\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} +\mathrm{8}^{\mathrm{2}} +\mathrm{13}^{\mathrm{2}} +\mathrm{21}^{\mathrm{2}} =\mathrm{21}×\mathrm{34}−\mathrm{1}=\mathrm{713} \\ $$
Commented by mr W last updated on 07/Jun/24
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Answered by A5T last updated on 07/Jun/24
Another proof(constructive): f_(n−1) f_(n+1) =(f_(n+1) −f_n )(f_n +f_(n−1) ) f_(n−1) f_(n+1) =f_n f_(n+1) +f_(n+1) f_(n−1) −f_n ^2 −f_n f_(n−1) ⇒f_n ^2 =f_n f_(n+1) −f_n f_(n−1) (we get a telescoping sum) ⇒f_1 ^2 +f_2 ^2 +...+f_n ^2 =f_n f_(n+1)
$${Another}\:{proof}\left({constructive}\right): \\ $$$${f}_{{n}−\mathrm{1}} {f}_{{n}+\mathrm{1}} =\left({f}_{{n}+\mathrm{1}} −{f}_{{n}} \right)\left({f}_{{n}} +{f}_{{n}−\mathrm{1}} \right) \\ $$$${f}_{{n}−\mathrm{1}} {f}_{{n}+\mathrm{1}} ={f}_{{n}} {f}_{{n}+\mathrm{1}} +{f}_{{n}+\mathrm{1}} {f}_{{n}−\mathrm{1}} −{f}_{{n}} ^{\mathrm{2}} −{f}_{{n}} {f}_{{n}−\mathrm{1}} \\ $$$$\Rightarrow{f}_{{n}} ^{\mathrm{2}} ={f}_{{n}} {f}_{{n}+\mathrm{1}} −{f}_{{n}} {f}_{{n}−\mathrm{1}} \:\left({we}\:{get}\:{a}\:{telescoping}\:{sum}\right) \\ $$$$\Rightarrow{f}_{\mathrm{1}} ^{\mathrm{2}} +{f}_{\mathrm{2}} ^{\mathrm{2}} +…+{f}_{{n}} ^{\mathrm{2}} ={f}_{{n}} {f}_{{n}+\mathrm{1}} \\ $$
Commented by mr W last updated on 07/Jun/24
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