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Question Number 72048 by ahmadshahhimat775@gmail.com last updated on 23/Oct/19

Answered by mind is power last updated on 23/Oct/19

$${\mathrm{a}}_2=\frac{6}{1+3}=\frac{3}{2}=\frac{3.2^0}{2}=\frac{2+1}{2}=\frac{2+2^0}{2}$$$${\mathrm{a}}_3=\frac{3}{1+\frac{3}{2}}=\frac{6}{5}=\frac{3.2}{3.2-1}=\frac{4+2}{4+2-1}=\frac{2+2^2}{2+2^2-1}$$$${\mathrm{a}}_4=\frac{\frac{12}{5}}{\frac{11}{5}}=\frac{12}{11}=\frac{3.2^2}{3.2^2-1}=\frac{2^3+2^2}{2^3+2^2-1}$$$${\mathrm{a}}_5=\frac{24}{23}=\frac{3.2^3}{3.2^3-1}=\frac{2^4+2^3}{2^4+2^3-1}$$$${\mathrm{a}}_{\mathrm{n}}=\frac{2^{\mathrm{n}-1}+2^{\mathrm{n}-2}}{2^{\mathrm{n}-1}+2^{\mathrm{n}-2}+1}\mathrm{if}\mathrm{n}⩾2$$$${\mathrm{a}}_1=3$$$$\mathrm{assum}{\mathrm{a}}_{\mathrm{n}}=\frac{2^{\mathrm{n}-1}+2^{\mathrm{n}-2}}{2^{\mathrm{n}-1}+2^{\mathrm{n}-2}-1},{\mathrm{a}}_{\mathrm{n}+1}=\frac{2^{\mathrm{n}+1-1}+2^{\mathrm{n}+1-2}}{2^{\mathrm{n}+1-1}+2^{\mathrm{n}+1-2}-1}?$$$${\mathrm{a}}_{\mathrm{n}+1}=\frac{2.\frac{2^{\mathrm{n}-1}+2^{\mathrm{n}-2}}{2^{\mathrm{n}-1}+2^{\mathrm{n}-2}-1}}{1+\frac{2^{\mathrm{n}-1}+2^{\mathrm{n}-2}}{2^{\mathrm{n}-1}+2^{\mathrm{n}-2}-1}}=\frac{\frac{2^{\mathrm{n}+1-1}+2^{\mathrm{n}+1-2}}{2^{\mathrm{n}-1}+2^{\mathrm{n}-2}-1}}{\frac{2^{\mathrm{n}-1}+2^{\mathrm{n}-1}+2^{\mathrm{n}-2}+2^{\mathrm{n}-2}-1}{2^{\mathrm{n}-1}+2^{\mathrm{n}-2}-1}}$$$$=\frac{2^{\mathrm{n}+1-1}+2^{\mathrm{n}+1-2}}{2^{\mathrm{n}+1-1}+2^{\mathrm{n}+1-2}-1}$$

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