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Question Number 57051 by Hassen_Timol last updated on 29/Mar/19

$$$$$${\mathit{F}}_{\mathit{n}}=\frac{{\phi }^n-{(1-\phi )}^n}{\sqrt{5}},\mathrm{with}\{\begin{array}{l}\phi =\frac{1+\sqrt{5}}{2}\end{array}$$$$$$$$\left({\mathit{G}}_{\mathit{n}}\right):\{\begin{array}{l}{\mathit{G}}_1=1\\ {\mathit{G}}_2=1\\ {\mathit{G}}_m={\mathit{G}}_{\mathrm{m}-1}+{\mathit{G}}_{m-2}\end{array}$$$$$$$$\mathrm{Give}\mathrm{a}\mathrm{proof}\mathrm{for}{\mathit{F}}_{\mathit{n}}={\mathit{G}}_{\mathit{n}},\mathrm{\forall }n\in {\mathbb{N}}^{\ast }.$$$$\mathrm{Thank}\mathrm{you}$$

Commented by prakash jain last updated on 29/Mar/19

$$G_m-G_{m-1}-G_{m-2}=0$$$$\mathrm{Characteristic}\mathrm{eqn}$$$${\lambda }^2-\lambda -1=0$$$$\lambda =\frac{1\pm \sqrt{5}}{2}$$$$G_n=c_1{\left(\frac{1+\sqrt{5}}{2}\right)}^{n-1}+c_2{\left(\frac{1-\sqrt{5}}{2}\right)}^{n-1}$$$$G_1=1\Rightarrow c_1+c_2=1$$$$G_2=1\Rightarrow c_1\left(\frac{1+\sqrt{5}}{2}\right)+c_2\left(\frac{1-\sqrt{5}}{2}\right)=1$$$$(1-c_2)\left(\frac{1+\sqrt{5}}{2}\right)+c_2\left(\frac{1-\sqrt{5}}{2}\right)=1$$$$\sqrt{5}{c}_2=\frac{\sqrt{5}-1}{2}$$$$c_2=\frac{\sqrt{5}-1}{2\sqrt{5}}$$$$c_1=\frac{\sqrt{5}+1}{2\sqrt{5}}$$$$G_n=\left(\frac{\sqrt{5}+1}{2\sqrt{5}}\right){\left(\frac{1+\sqrt{5}}{2}\right)}^{n-1}+\left(\frac{\sqrt{5}-1}{2\sqrt{5}}\right){\left(\frac{1-\sqrt{5}}{2}\right)}^{n-1}$$$$=\frac{{\left(\frac{1+\sqrt{5}}{2}\right)}^n}{\sqrt{5}}-\frac{{\left(\frac{1-\sqrt{5}}{2}\right)}^n}{\sqrt{5}}$$$$\phi =\frac{1+\sqrt{5}}{2},1-\phi =\frac{1-\sqrt{5}}{2}$$$$G_n=\frac{{\phi }^n}{\sqrt{5}}-\frac{{(1-\phi )}^n}{\sqrt{5}}=\frac{{\phi }^n-{(1-\phi )}^n}{\sqrt{5}}=F_n$$$$$$

Commented by Hassen_Timol last updated on 29/Mar/19

$$\mathrm{Thank}\mathrm{you}\mathrm{for}\mathrm{the}\mathrm{answer},\mathrm{sir}!$$

Commented by Hassen_Timol last updated on 29/Mar/19

$$\mathrm{Could}\mathrm{you}\mathrm{explain},\mathrm{please},\mathrm{a}\mathrm{little}\mathrm{bit}$$$$\mathrm{more}\mathrm{in}\mathrm{details}?\mathrm{it}\mathrm{would}\mathrm{be}\mathrm{helpful}...$$

Commented by mr W last updated on 29/Mar/19

$$seeQ56864,maybeitcanhelpyouto$$$$understandhowtofindexplictformula$$$$for{G}_m.$$

Commented by Hassen_Timol last updated on 30/Mar/19

$$\mathrm{Thank}\mathrm{you}$$

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