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Question Number 165361 by MikeH last updated on 31/Jan/22

Obtain a general formula for the sequence (2/3),(4/5),(8/9),((16)/(17)),((32)/(33)),... assuming the sequence continues in that pattern.

$$\mathrm{Obtain}\:\mathrm{a}\:\mathrm{general}\:\mathrm{formula}\:\mathrm{for} \\ $$$$\mathrm{the}\:\mathrm{sequence} \\ $$$$\:\frac{\mathrm{2}}{\mathrm{3}},\frac{\mathrm{4}}{\mathrm{5}},\frac{\mathrm{8}}{\mathrm{9}},\frac{\mathrm{16}}{\mathrm{17}},\frac{\mathrm{32}}{\mathrm{33}},... \\ $$$$\mathrm{assuming}\:\mathrm{the}\:\mathrm{sequence}\:\mathrm{continues}\:\mathrm{in}\:\mathrm{that} \\ $$$$\mathrm{pattern}. \\ $$

Commented by MJS_new last updated on 31/Jan/22

((7n^4 )/(33660))−((13n^3 )/(16830))−((767n^2 )/(33660))+((3433n)/(16830))+((818)/(1683))

$$\frac{\mathrm{7}{n}^{\mathrm{4}} }{\mathrm{33660}}−\frac{\mathrm{13}{n}^{\mathrm{3}} }{\mathrm{16830}}−\frac{\mathrm{767}{n}^{\mathrm{2}} }{\mathrm{33660}}+\frac{\mathrm{3433}{n}}{\mathrm{16830}}+\frac{\mathrm{818}}{\mathrm{1683}} \\ $$

Commented by MJS_new last updated on 31/Jan/22

you can like or dislike it but it still is one of zillions of right answers

$$\mathrm{you}\:\mathrm{can}\:\mathrm{like}\:\mathrm{or}\:\mathrm{dislike}\:\mathrm{it}\:\mathrm{but}\:\mathrm{it}\:\mathrm{still}\:\mathrm{is}\:\mathrm{one} \\ $$$$\mathrm{of}\:\mathrm{zillions}\:\mathrm{of}\:\mathrm{right}\:\mathrm{answers} \\ $$

Answered by Rasheed.Sindhi last updated on 31/Jan/22

(2/3),(4/5),(8/9),((16)/(17)),((32)/(33)),... (2/(2+1)),(4/(4+1)),(8/(8+1)),((16)/(16+1)),((32)/(32+1)),... (2^1 /(2^1 +1)),(2^2 /(2^2 +1)),(2^3 /(2^3 +1)),(2^4 /(2^4 +1)),(2^5 /(2^5 +1)),...,(2^n /(2^n +1))

$$\:\frac{\mathrm{2}}{\mathrm{3}},\frac{\mathrm{4}}{\mathrm{5}},\frac{\mathrm{8}}{\mathrm{9}},\frac{\mathrm{16}}{\mathrm{17}},\frac{\mathrm{32}}{\mathrm{33}},... \\ $$$$\frac{\mathrm{2}}{\mathrm{2}+\mathrm{1}},\frac{\mathrm{4}}{\mathrm{4}+\mathrm{1}},\frac{\mathrm{8}}{\mathrm{8}+\mathrm{1}},\frac{\mathrm{16}}{\mathrm{16}+\mathrm{1}},\frac{\mathrm{32}}{\mathrm{32}+\mathrm{1}},... \\ $$$$\frac{\mathrm{2}^{\mathrm{1}} }{\mathrm{2}^{\mathrm{1}} +\mathrm{1}},\frac{\mathrm{2}^{\mathrm{2}} }{\mathrm{2}^{\mathrm{2}} +\mathrm{1}},\frac{\mathrm{2}^{\mathrm{3}} }{\mathrm{2}^{\mathrm{3}} +\mathrm{1}},\frac{\mathrm{2}^{\mathrm{4}} }{\mathrm{2}^{\mathrm{4}} +\mathrm{1}},\frac{\mathrm{2}^{\mathrm{5}} }{\mathrm{2}^{\mathrm{5}} +\mathrm{1}},...,\frac{\mathrm{2}^{{n}} }{\mathrm{2}^{{n}} +\mathrm{1}} \\ $$

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