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Question Number 81446 by john santu last updated on 13/Feb/20

$$\mathrm{given}{\mathrm{a}}_1=2,{\mathrm{a}}_2=3\mathrm{and}$$$${\mathrm{a}}_{\mathrm{n}+2}={\mathrm{a}}_{\mathrm{n}+1}+\frac{{\mathrm{a}}_{\mathrm{n}}}{2}$$$$\mathrm{find}{\mathrm{a}}_{\mathrm{n}}=?$$

Commented by mr W last updated on 13/Feb/20

$$q^2-q-\frac{1}{2}=0$$$$q=\frac{1\pm \sqrt{3}}{2}$$$$a_n=A{\left(\frac{1+\sqrt{3}}{2}\right)}^n+B{\left(\frac{1-\sqrt{3}}{2}\right)}^n$$$$a_1=A\left(\frac{1+\sqrt{3}}{2}\right)+B\left(\frac{1-\sqrt{3}}{2}\right)=2$$$$a_2=A{\left(\frac{1+\sqrt{3}}{2}\right)}^2+B{\left(\frac{1-\sqrt{3}}{2}\right)}^2=3$$$$\Rightarrow A+B=2$$$$\Rightarrow A-B=\frac{2\sqrt{3}}{3}$$$$\Rightarrow A=\frac{3+\sqrt{3}}{3}$$$$\Rightarrow B=\frac{3-\sqrt{3}}{3}$$$$\Rightarrow a_n=\left(\frac{3+\sqrt{3}}{3}\right){\left(\frac{1+\sqrt{3}}{2}\right)}^n+\left(\frac{3-\sqrt{3}}{3}\right){\left(\frac{1-\sqrt{3}}{2}\right)}^n$$

Commented by behi83417@gmail.com last updated on 13/Feb/20

$${\mathrm{a}}_{\mathrm{n}}=\frac{1}{\sqrt{3}}[{\left(\frac{1+\sqrt{3}}{2}\right)}^{\mathrm{n}+1}-{\left(\frac{1-\sqrt{3}}{2}\right)}^{\mathrm{n}+1}]$$

Commented by jagoll last updated on 13/Feb/20

$$\mathrm{from}{\mathrm{a}}_{\mathrm{n}}=\left(\frac{3+\sqrt{3}}{3}\right){\left(\frac{1+\sqrt{3}}{2}\right)}^{\mathrm{n}}+\left(\frac{3-\sqrt{3}}{3}\right){\left(\frac{1-\sqrt{3}}{2}\right)}^{\mathrm{n}}$$$${\mathrm{a}}_1=\left(\frac{3+\sqrt{3}}{3}\right)\left(\frac{1+\sqrt{3}}{2}\right)+\left(\frac{3-\sqrt{3}}{3}\right)\left(\frac{1-\sqrt{3}}{2}\right)$$$$=\frac{6+4\sqrt{3}+6-4\sqrt{3}}{6}=\frac{12}{6}=2$$$$\mathrm{amazing}\mathrm{sir}$$

Commented by john santu last updated on 13/Feb/20

$$\mathrm{in}\mathrm{generally}$$$$\mathrm{if}\Rightarrow {2\mathrm{a}}_{\mathrm{n}+2}={3\mathrm{a}}_{\mathrm{n}+1}-\frac{{\mathrm{a}}_{\mathrm{n}}}{3}$$$$\Rightarrow {2\mathrm{q}}^2-3\mathrm{q}+\frac{1}{3}=0$$$${6\mathrm{q}}^2-9\mathrm{q}+1=0$$$$\mathrm{q}=\frac{9\pm \sqrt{81-24}}{12}=\frac{9\pm \sqrt{57}}{12}$$$${\mathrm{a}}_{\mathrm{n}}=\mathrm{A}{\left(\frac{9-\sqrt{57}}{12}\right)}^{\mathrm{n}}+\mathrm{B}{\left(\frac{9+\sqrt{57}}{12}\right)}^{\mathrm{n}}$$

Commented by jagoll last updated on 13/Feb/20

$$\mathrm{thank}\mathrm{you}\mathrm{mister}\mathrm{W}.\mathrm{i}\mathrm{have}\mathrm{the}$$$$\mathrm{same}\mathrm{problem}\mathrm{with}\mathrm{this}\mathrm{kind}.$$$$$$

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