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Question Number 111536 by Aina Samuel Temidayo last updated on 04/Sep/20

Let 2,3,5,6,7,10,11,... be increasing sequence of positive integers that are neither the square nor cube of an integer. Find the 2016th term of this sequence.

$$\mathrm{Let}\:\mathrm{2},\mathrm{3},\mathrm{5},\mathrm{6},\mathrm{7},\mathrm{10},\mathrm{11},...\:\mathrm{be}\:\mathrm{increasing} \\ $$$$\mathrm{sequence}\:\mathrm{of}\:\mathrm{positive}\:\mathrm{integers}\:\mathrm{that}\:\mathrm{are} \\ $$$$\mathrm{neither}\:\mathrm{the}\:\mathrm{square}\:\mathrm{nor}\:\mathrm{cube}\:\mathrm{of}\:\mathrm{an} \\ $$$$\mathrm{integer}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{2016th}\:\mathrm{term}\:\mathrm{of}\:\mathrm{this} \\ $$$$\mathrm{sequence}. \\ $$

Commented by Rasheed.Sindhi last updated on 04/Sep/20

2070

$$\mathrm{2070} \\ $$

Commented by Aina Samuel Temidayo last updated on 04/Sep/20

Your solution please?

$$\mathrm{Your}\:\mathrm{solution}\:\mathrm{please}? \\ $$

Answered by Lordose last updated on 04/Sep/20

The sequence is an A.P U_n =a+(n−1)d a=2, d=1 n=2016 U_n =2+2015=2017

$$\boldsymbol{\mathrm{The}}\:\boldsymbol{\mathrm{sequence}}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{an}}\:\boldsymbol{\mathrm{A}}.\boldsymbol{\mathrm{P}} \\ $$$$\boldsymbol{\mathrm{U}}_{\mathrm{n}} =\boldsymbol{\mathrm{a}}+\left(\boldsymbol{\mathrm{n}}−\mathrm{1}\right)\boldsymbol{\mathrm{d}} \\ $$$$\boldsymbol{\mathrm{a}}=\mathrm{2},\:\boldsymbol{\mathrm{d}}=\mathrm{1}\:\boldsymbol{\mathrm{n}}=\mathrm{2016} \\ $$$$\boldsymbol{\mathrm{U}}_{\boldsymbol{\mathrm{n}}} =\mathrm{2}+\mathrm{2015}=\mathrm{2017} \\ $$

Commented by Aina Samuel Temidayo last updated on 04/Sep/20

How is it an A.P? The common difference is not uniform.

$$\mathrm{How}\:\mathrm{is}\:\mathrm{it}\:\mathrm{an}\:\mathrm{A}.\mathrm{P}?\:\mathrm{The}\:\mathrm{common} \\ $$$$\mathrm{difference}\:\mathrm{is}\:\mathrm{not}\:\mathrm{uniform}. \\ $$

Answered by Rasheed.Sindhi last updated on 06/Sep/20

∣{2,3,5,6,7,10,11,...(2016 members)}∣ =∣{1,2,3,...,n}∣ −∣{1,4,9,...,(⌊(√n)⌋)^2 }∣ −∣{1,8,27,...,(⌊(n)^(1/3) ⌋)^3 }∣ +∣{1,64,729,..(⌊(n)^(1/6) ⌋)^6 }∣ 2016=n−⌊(√n)⌋−⌊(n)^(1/3) ⌋+⌊(n)^(1/6) ⌋ n=2070 satisfy the above equation Hence 2016th term=2070

$$\mid\left\{\mathrm{2},\mathrm{3},\mathrm{5},\mathrm{6},\mathrm{7},\mathrm{10},\mathrm{11},...\left(\mathrm{2016}\:{members}\right)\right\}\mid \\ $$$$\:\:\:\:\:\:=\mid\left\{\mathrm{1},\mathrm{2},\mathrm{3},...,{n}\right\}\mid \\ $$$$\:\:\:\:\:\:\:\:\:\:\:−\mid\left\{\mathrm{1},\mathrm{4},\mathrm{9},...,\left(\lfloor\sqrt{{n}}\rfloor\right)^{\mathrm{2}} \right\}\mid \\ $$$$\:\:\:\:\:\:\:\:\:\:\:−\mid\left\{\mathrm{1},\mathrm{8},\mathrm{27},...,\left(\lfloor\sqrt[{\mathrm{3}}]{{n}}\rfloor\right)^{\mathrm{3}} \right\}\mid \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:+\mid\left\{\mathrm{1},\mathrm{64},\mathrm{729},..\left(\lfloor\sqrt[{\mathrm{6}}]{{n}}\rfloor\right)^{\mathrm{6}} \right\}\mid \\ $$$$\mathrm{2016}={n}−\lfloor\sqrt{{n}}\rfloor−\lfloor\sqrt[{\mathrm{3}}]{{n}}\rfloor+\lfloor\sqrt[{\mathrm{6}}]{{n}}\rfloor \\ $$$${n}=\mathrm{2070}\:{satisfy}\:{the}\:{above}\:{equation} \\ $$$${Hence}\:\mathrm{2016}{th}\:{term}=\mathrm{2070} \\ $$

Commented by Aina Samuel Temidayo last updated on 06/Sep/20

Thanks.

$$\mathrm{Thanks}. \\ $$

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