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If-n-2-and-U-n-3-5-n-3-5-n-then-prove-that-U-n-1-6U-n-4U-n-1-




Question Number 202328 by MATHEMATICSAM last updated on 24/Dec/23
If n ≥ 2 and U_n  = (3 + (√5))^n  + (3 − (√5))^n   then prove that U_(n + 1)  = 6U_n  − 4U_(n − 1)  .
$$\mathrm{If}\:{n}\:\geqslant\:\mathrm{2}\:\mathrm{and}\:\mathrm{U}_{{n}} \:=\:\left(\mathrm{3}\:+\:\sqrt{\mathrm{5}}\right)^{{n}} \:+\:\left(\mathrm{3}\:−\:\sqrt{\mathrm{5}}\right)^{{n}} \\ $$$$\mathrm{then}\:\mathrm{prove}\:\mathrm{that}\:\mathrm{U}_{{n}\:+\:\mathrm{1}} \:=\:\mathrm{6U}_{{n}} \:−\:\mathrm{4U}_{{n}\:−\:\mathrm{1}} \:. \\ $$
Commented by aleks041103 last updated on 24/Dec/23
It is true even for n=1
$${It}\:{is}\:{true}\:{even}\:{for}\:{n}=\mathrm{1} \\ $$
Answered by aleks041103 last updated on 24/Dec/23
U_(n+1) +4U_(n−1) −6U_n =  =(3+(√5))^(n−1) ((3+(√5))^2 +4−6(3+(√5)))+  (3−(√5))^(n−1) ((3−(√5))^2 +4−6(3−(√5)))  (3+(√5))^2 =14+6(√5)  ⇒(14+6(√5))+4−18−6(√5)=0  (3−(√5))^2 =14−6(√5)  (14−6(√5))+4−18+6(√5)=0  ⇒U_(n+1) −6U_n +4U_(n−1) =  =(3+(√5))^(n−1) (0)+(3−(√5))^(n−1) (0)=0    ⇒U_(n+1) =6U_n −4U_(n−1)
$${U}_{{n}+\mathrm{1}} +\mathrm{4}{U}_{{n}−\mathrm{1}} −\mathrm{6}{U}_{{n}} = \\ $$$$=\left(\mathrm{3}+\sqrt{\mathrm{5}}\right)^{{n}−\mathrm{1}} \left(\left(\mathrm{3}+\sqrt{\mathrm{5}}\right)^{\mathrm{2}} +\mathrm{4}−\mathrm{6}\left(\mathrm{3}+\sqrt{\mathrm{5}}\right)\right)+ \\ $$$$\left(\mathrm{3}−\sqrt{\mathrm{5}}\right)^{{n}−\mathrm{1}} \left(\left(\mathrm{3}−\sqrt{\mathrm{5}}\right)^{\mathrm{2}} +\mathrm{4}−\mathrm{6}\left(\mathrm{3}−\sqrt{\mathrm{5}}\right)\right) \\ $$$$\left(\mathrm{3}+\sqrt{\mathrm{5}}\right)^{\mathrm{2}} =\mathrm{14}+\mathrm{6}\sqrt{\mathrm{5}} \\ $$$$\Rightarrow\left(\mathrm{14}+\mathrm{6}\sqrt{\mathrm{5}}\right)+\mathrm{4}−\mathrm{18}−\mathrm{6}\sqrt{\mathrm{5}}=\mathrm{0} \\ $$$$\left(\mathrm{3}−\sqrt{\mathrm{5}}\right)^{\mathrm{2}} =\mathrm{14}−\mathrm{6}\sqrt{\mathrm{5}} \\ $$$$\left(\mathrm{14}−\mathrm{6}\sqrt{\mathrm{5}}\right)+\mathrm{4}−\mathrm{18}+\mathrm{6}\sqrt{\mathrm{5}}=\mathrm{0} \\ $$$$\Rightarrow{U}_{{n}+\mathrm{1}} −\mathrm{6}{U}_{{n}} +\mathrm{4}{U}_{{n}−\mathrm{1}} = \\ $$$$=\left(\mathrm{3}+\sqrt{\mathrm{5}}\right)^{{n}−\mathrm{1}} \left(\mathrm{0}\right)+\left(\mathrm{3}−\sqrt{\mathrm{5}}\right)^{{n}−\mathrm{1}} \left(\mathrm{0}\right)=\mathrm{0} \\ $$$$ \\ $$$$\Rightarrow{U}_{{n}+\mathrm{1}} =\mathrm{6}{U}_{{n}} −\mathrm{4}{U}_{{n}−\mathrm{1}} \\ $$

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