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Question Number 157343 by mathocean1 last updated on 22/Oct/21

Answered by mindispower last updated on 23/Oct/21

X^2 −X−1=0 ⇒X∈{((1+(√5))/2),((1−(√5))/2)} u_n =a(((1+(√5))/2))^n +b(((1−(√5))/2))^n (a,b) { ((a+b=0)),(((1/2)(a+b)+((√5)/2)(a−b)=1)) :} ⇒a=(1/( (√5))),b=−(1/( (√5))) ∅_n =(1/( (√5)))((((1+(√5))/2))^n −(((1−(√5))/2))^n ) (2)∅_(n+1) ^2 =(1/5)((((1+(√5))/2))^(2n+2) +(((1−(√5))/2))^(2n+2) −2(−1)^(n+1) ) ∅_n .∅_(n+2) =(1/5)((((1+(√5))/2))^(2n+2) +(((1−(√5))/2))^(2n+2) (−1)^n (−(((1+(√5))/2))^2 −(((1−(√5))/2))^2 ) =(1/5)((((1+(√5))/2))^(2+2n) +(((1−(√5))/2))^(2n+2) +(−1)^n (−3)) ∅_(n+1) ^2 −∅_n ∅_(n+1) =(−1)^n (3Σ)1−((∅_n /∅_(n+1) ))=(((−1)^n )/((∅_(n+1) )^2 )) l=lim_(n→∞) ((∅_(n+1) /∅_n )) 1−(1/l)=0⇒l=1 (4),Σ_(k=0) ^n C_n ^k ∅_k =(1/( (√5)))(Σ_(k=0) ^n C_n ^k (((1+(√5))/2))^k −5)Σ_(k≤n) C_n ^k (((1−(√5))/2))^(k)) =(1/( (√5)))((1+((1+(√5))/2))^n −(1+((1−(√5))/2))^n ) =(1/( (√5)))((((6+2(√5))/4))^n −(((6−2(√5))/4))^n )=(1/( (√5)))((((1+(√5))/2))^(2n) −(((1−(√5))/2))^(2n) )=∅_(2n) (((1+_− (√5))/2))^2 =((6+_− 2(√5))/4) (5) meme Chose Que la d1

$${X}^{\mathrm{2}} −{X}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{X}\in\left\{\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}},\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right\} \\ $$$${u}_{{n}} ={a}\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} +{b}\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} \\ $$$$\left({a},{b}\right)\begin{cases}{{a}+{b}=\mathrm{0}}\\{\frac{\mathrm{1}}{\mathrm{2}}\left({a}+{b}\right)+\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\left({a}−{b}\right)=\mathrm{1}}\end{cases} \\ $$$$\Rightarrow{a}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}},{b}=−\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}} \\ $$$$\emptyset_{{n}} =\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left(\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} −\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} \right) \\ $$$$\left(\mathrm{2}\right)\emptyset_{{n}+\mathrm{1}} ^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{5}}\left(\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{2}{n}+\mathrm{2}} +\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{2}{n}+\mathrm{2}} −\mathrm{2}\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \right) \\ $$$$\emptyset_{{n}} .\emptyset_{{n}+\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{5}}\left(\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{2}{n}+\mathrm{2}} +\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{2}{n}+\mathrm{2}} \left(−\mathrm{1}\right)^{{n}} \left(−\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{2}} −\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{2}} \right)\right. \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$=\frac{\mathrm{1}}{\mathrm{5}}\left(\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{2}+\mathrm{2}{n}} +\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{2}{n}+\mathrm{2}} +\left(−\mathrm{1}\right)^{{n}} \left(−\mathrm{3}\right)\right) \\ $$$$\emptyset_{{n}+\mathrm{1}} ^{\mathrm{2}} −\emptyset_{{n}} \emptyset_{{n}+\mathrm{1}} =\left(−\mathrm{1}\right)^{{n}} \\ $$$$\left(\mathrm{3}\Sigma\right)\mathrm{1}−\left(\frac{\emptyset_{{n}} }{\emptyset_{{n}+\mathrm{1}} }\right)=\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\emptyset_{{n}+\mathrm{1}} \right)^{\mathrm{2}} } \\ $$$${l}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\emptyset_{{n}+\mathrm{1}} }{\emptyset_{{n}} }\right) \\ $$$$\mathrm{1}−\frac{\mathrm{1}}{{l}}=\mathrm{0}\Rightarrow{l}=\mathrm{1} \\ $$$$\left(\mathrm{4}\right),\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{C}_{{n}} ^{{k}} \emptyset_{{k}} =\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left(\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{C}_{{n}} ^{{k}} \left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{k}} −\mathrm{5}\right)\underset{{k}\leqslant{n}} {\sum}{C}_{{n}} ^{{k}} \left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\left.{k}\right)} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left(\left(\mathrm{1}+\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} −\left(\mathrm{1}+\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} \right) \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left(\left(\frac{\mathrm{6}+\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{4}}\right)^{{n}} −\left(\frac{\mathrm{6}−\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{4}}\right)^{{n}} \right)=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left(\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{2}{n}} −\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{2}{n}} \right)=\emptyset_{\mathrm{2}{n}} \\ $$$$\left(\frac{\mathrm{1}\underset{−} {+}\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{\mathrm{6}\underset{−} {+}\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{4}} \\ $$$$\left(\mathrm{5}\right)\:{meme}\:{Chose}\:{Que}\:{la}\:\:{d}\mathrm{1} \\ $$$$ \\ $$

Commented by mathocean1 last updated on 24/Oct/21

Thanks

$${Thanks} \\ $$

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