Question Number 222105 by alcohol last updated on 17/Jun/25
Commented by alcohol last updated on 17/Jun/25
$${please}\:{help}\:{with}\:\left({iii}\right)\:{a}\:{and}\:{b}\: \\ $$
Answered by mr W last updated on 19/Jun/25
![F_(n+2) −F_(n+1) −F_n =0 r^2 −r−1=0 r_1 =((1+(√5))/2), r_2 =((1−(√5))/2) F_n =A(((1+(√5))/2))^n +B(((1−(√5))/2))^n F_0 =0 ⇒A+B=0 ⇒B=−A F_1 =A(((1+(√5))/2))+B(((1−(√5))/2))=1 ⇒A=(1/( (√5))) ⇒F_n =(1/( (√5)))(((1+(√5))/2))^n −(1/( (√5)))(((1−(√5))/2))^n (a) F_(2n) =(1/( (√5)))(((1+(√5))/2))^(2n) −(1/( (√5)))(((1−(√5))/2))^(2n) F_(2n) =(1/( (√5)))(1+((1+(√5))/2))^n −(1/( (√5)))(1+((1−(√5))/2))^n F_(2n) =(1/( (√5)))Σ_(k=0) ^n ((n),(k) )(((1+(√5))/2))^k −(1/( (√5)))Σ_(k=0) ^n ((n),(k) )(((1−(√5))/2))^k F_(2n) =Σ_(k=0) ^n ((n),(k) )[(1/( (√5)))(((1+(√5))/2))^k −(1/( (√5)))(((1−(√5))/2))^k ] F_(2n) =Σ_(k=0) ^n ((n),(k) )F_k ✓ (b.1) F_(3n) =(1/( (√5)))(((1+(√5))/2))^(3n) −(1/( (√5)))(((1−(√5))/2))^(3n) F_(3n) =(1/( (√5)))(2+(√5))^n −(1/( (√5)))(2−(√5))^n F_(3n) =(1/( (√5)))(1+2×((1+(√5))/2))^n −(1/( (√5)))(1+2×((1−(√5))/2))^n F_(3n) =(1/( (√5)))Σ_(k=0) ^n [ ((n),(k) )2^k ×(((1+(√5))/2))^k ]−(1/( (√5)))Σ_(k=0) ^n [ ((n),(k) )2^k ×(((1−(√5))/2))^k ] F_(3n) =Σ_(k=0) ^n ((n),(k) )2^k [(1/( (√5)))(((1+(√5))/2))^k −(1/( (√5)))(((1−(√5))/2))^k ] F_(3n) =Σ_(k=0) ^n ((n),(k) )2^k F_k ✓ (b.2) F_(3n) =(1/( (√5)))(((1+(√5))/2))^n (((1+(√5))/2))^(2n) −(1/( (√5)))(((1−(√5))/2))^n (((1−(√5))/2))^(2n) F_(3n) =(1/( (√5)))(((1+(√5))/2))^n (1+((1+(√5))/2))^n −(1/( (√5)))(((1−(√5))/2))^n (1+((1−(√5))/2))^n F_(3n) =(1/( (√5)))(((1+(√5))/2))^n Σ_(k=0) ^n ((n),(k) )(((1+(√5))/2))^k −(1/( (√5)))(((1−(√5))/2))^n Σ_(k=0) ^n ((n),(k) )(((1−(√5))/2))^k F_(3n) =Σ_(k=0) ^n ((n),(k) )[(1/( (√5)))(((1+(√5))/2))^(n+k) −(1/( (√5)))(((1−(√5))/2))^(n+k) ] F_(3n) =Σ_(k=0) ^n ((n),(k) )F_(n+k) ✓](https://www.tinkutara.com/question/Q222150.png)
$${F}_{{n}+\mathrm{2}} −{F}_{{n}+\mathrm{1}} −{F}_{{n}} =\mathrm{0} \\ $$$${r}^{\mathrm{2}} −{r}−\mathrm{1}=\mathrm{0} \\ $$$${r}_{\mathrm{1}} =\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}},\:{r}_{\mathrm{2}} =\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$${F}_{{n}} ={A}\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} +{B}\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} \\ $$$${F}_{\mathrm{0}} =\mathrm{0}\:\Rightarrow{A}+{B}=\mathrm{0}\:\Rightarrow{B}=−{A} \\ $$$${F}_{\mathrm{1}} ={A}\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)+{B}\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)=\mathrm{1}\:\Rightarrow{A}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}} \\ $$$$\Rightarrow{F}_{{n}} =\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} −\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} \\ $$$$\left({a}\right) \\ $$$${F}_{\mathrm{2}{n}} =\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{2}{n}} −\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{2}{n}} \\ $$$${F}_{\mathrm{2}{n}} =\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left(\mathrm{1}+\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} −\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left(\mathrm{1}+\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} \\ $$$${F}_{\mathrm{2}{n}} =\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{k}} −\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{k}} \\ $$$${F}_{\mathrm{2}{n}} =\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}\left[\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{k}} −\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{k}} \right] \\ $$$${F}_{\mathrm{2}{n}} =\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}{F}_{{k}} \:\:\checkmark \\ $$$$\left({b}.\mathrm{1}\right) \\ $$$${F}_{\mathrm{3}{n}} =\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{3}{n}} −\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{3}{n}} \\ $$$${F}_{\mathrm{3}{n}} =\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)^{{n}} −\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left(\mathrm{2}−\sqrt{\mathrm{5}}\right)^{{n}} \\ $$$${F}_{\mathrm{3}{n}} =\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left(\mathrm{1}+\mathrm{2}×\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} −\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left(\mathrm{1}+\mathrm{2}×\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} \\ $$$${F}_{\mathrm{3}{n}} =\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\left[\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}\mathrm{2}^{{k}} ×\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{k}} \right]−\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\left[\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}\mathrm{2}^{{k}} ×\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{k}} \right] \\ $$$${F}_{\mathrm{3}{n}} =\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}\mathrm{2}^{{k}} \left[\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{k}} −\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{k}} \right] \\ $$$${F}_{\mathrm{3}{n}} =\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}\mathrm{2}^{{k}} {F}_{{k}} \:\:\checkmark \\ $$$$\left({b}.\mathrm{2}\right) \\ $$$${F}_{\mathrm{3}{n}} =\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} \left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{2}{n}} −\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} \left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{2}{n}} \\ $$$${F}_{\mathrm{3}{n}} =\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} \left(\mathrm{1}+\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} −\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} \left(\mathrm{1}+\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} \\ $$$${F}_{\mathrm{3}{n}} =\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} \underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{k}} −\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} \underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{k}} \\ $$$${F}_{\mathrm{3}{n}} =\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}\left[\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}+{k}} −\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}+{k}} \right] \\ $$$${F}_{\mathrm{3}{n}} =\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}{F}_{{n}+{k}} \:\:\:\checkmark \\ $$