Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 118936 by mathocean1 last updated on 20/Oct/20

show by recurrence that: for n ∈ N^∗ , 2^(6n−5) +3^(2n ) is divisible by 11.

$${show}\:{by}\:{recurrence}\:{that}: \\ $$$${for}\:{n}\:\in\:\mathbb{N}^{\ast} ,\:\mathrm{2}^{\mathrm{6}{n}−\mathrm{5}} +\mathrm{3}^{\mathrm{2}{n}\:} \:{is}\:{divisible}\:{by} \\ $$$$\mathrm{11}. \\ $$

Answered by Dwaipayan Shikari last updated on 20/Oct/20

2^(6n−5) +3^(2n) =11d=Ψ(n) 2^(6n+1) +9^(n+1) =Ψ(n+1) (Replacing n as n+1) =2^(6n+1) +2.2^(6n−5) +9^(n+1) +2.9^n −2(2^(6n−5) +9^n ) =2^(6n−5) (2^6 +2)+9^n (9+2)−22d (As Ψ(n)=2^(6n−5) +3^(2n) =11d) =66.2^(6n−5) +11.9^n −22d Which is divisible by 11 Both Ψ(n) and Ψ(n+1) is divisible by 11 So Generally Ψ(n)=2^(6n−5) +3^(2n) is divisble by 11

$$\mathrm{2}^{\mathrm{6}{n}−\mathrm{5}} +\mathrm{3}^{\mathrm{2}{n}} =\mathrm{11}{d}=\Psi\left({n}\right) \\ $$$$\mathrm{2}^{\mathrm{6}{n}+\mathrm{1}} +\mathrm{9}^{{n}+\mathrm{1}} =\Psi\left({n}+\mathrm{1}\right)\:\:\:\:\:\:\:\:\left({Replacing}\:{n}\:{as}\:{n}+\mathrm{1}\right) \\ $$$$=\mathrm{2}^{\mathrm{6}{n}+\mathrm{1}} +\mathrm{2}.\mathrm{2}^{\mathrm{6}{n}−\mathrm{5}} +\mathrm{9}^{{n}+\mathrm{1}} +\mathrm{2}.\mathrm{9}^{{n}} −\mathrm{2}\left(\mathrm{2}^{\mathrm{6}{n}−\mathrm{5}} +\mathrm{9}^{{n}} \right) \\ $$$$=\mathrm{2}^{\mathrm{6}{n}−\mathrm{5}} \left(\mathrm{2}^{\mathrm{6}} +\mathrm{2}\right)+\mathrm{9}^{{n}} \left(\mathrm{9}+\mathrm{2}\right)−\mathrm{22}{d}\:\:\:\:\left({As}\:\Psi\left({n}\right)=\mathrm{2}^{\mathrm{6}{n}−\mathrm{5}} +\mathrm{3}^{\mathrm{2}{n}} =\mathrm{11}{d}\right) \\ $$$$=\mathrm{66}.\mathrm{2}^{\mathrm{6}{n}−\mathrm{5}} +\mathrm{11}.\mathrm{9}^{{n}} −\mathrm{22}{d} \\ $$$${Which}\:{is}\:{divisible}\:{by}\:\mathrm{11} \\ $$$${Both}\:\Psi\left({n}\right)\:{and}\:\Psi\left({n}+\mathrm{1}\right)\:{is}\:{divisible}\:{by}\:\mathrm{11} \\ $$$${So} \\ $$$${Generally}\:\Psi\left({n}\right)=\mathrm{2}^{\mathrm{6}{n}−\mathrm{5}} +\mathrm{3}^{\mathrm{2}{n}} \:\:{is}\:{divisble}\:{by}\:\mathrm{11} \\ $$

Answered by Bird last updated on 21/Oct/20

let u_n =2^(6n−5 ) +3^(2n) n=1 ⇒u_1 =2+3^2 =11 this number is divisible by 11 let suppose u_n divisible by 11 we have u_(n+1) =2^(6n+6−5) +3^(2n+2) =2^6 (u_n −3^(2n) )+3^(2n) ×9 =2^6 u_n −2^6 3^(2n) +3^(2n) ×9 =2^6 u_n +3^(2n) (9−2^6 ) =2^6 (11k)+55.3^(2n) =11{2^6 k+5.3^(2n) } ⇒u_(n+1) is divisible by 11 the telation is true at term n+1

$${let}\:{u}_{{n}} =\mathrm{2}^{\mathrm{6}{n}−\mathrm{5}\:} +\mathrm{3}^{\mathrm{2}{n}} \\ $$$${n}=\mathrm{1}\:\Rightarrow{u}_{\mathrm{1}} =\mathrm{2}+\mathrm{3}^{\mathrm{2}} =\mathrm{11}\:{this}\:{number} \\ $$$${is}\:{divisible}\:{by}\:\mathrm{11}\:\: \\ $$$${let}\:{suppose}\:{u}_{{n}} {divisible}\:{by}\:\mathrm{11}\: \\ $$$${we}\:{have}\:{u}_{{n}+\mathrm{1}} =\mathrm{2}^{\mathrm{6}{n}+\mathrm{6}−\mathrm{5}} +\mathrm{3}^{\mathrm{2}{n}+\mathrm{2}} \\ $$$$=\mathrm{2}^{\mathrm{6}} \left({u}_{{n}} −\mathrm{3}^{\mathrm{2}{n}} \right)+\mathrm{3}^{\mathrm{2}{n}} ×\mathrm{9} \\ $$$$=\mathrm{2}^{\mathrm{6}} {u}_{{n}} −\mathrm{2}^{\mathrm{6}} \:\mathrm{3}^{\mathrm{2}{n}} +\mathrm{3}^{\mathrm{2}{n}} \:×\mathrm{9} \\ $$$$=\mathrm{2}^{\mathrm{6}} {u}_{{n}} +\mathrm{3}^{\mathrm{2}{n}} \left(\mathrm{9}−\mathrm{2}^{\mathrm{6}} \right) \\ $$$$=\mathrm{2}^{\mathrm{6}} \left(\mathrm{11}{k}\right)+\mathrm{55}.\mathrm{3}^{\mathrm{2}{n}} \\ $$$$=\mathrm{11}\left\{\mathrm{2}^{\mathrm{6}} {k}+\mathrm{5}.\mathrm{3}^{\mathrm{2}{n}} \right\}\:\Rightarrow{u}_{{n}+\mathrm{1}} {is} \\ $$$${divisible}\:{by}\:\mathrm{11}\:{the}\:{telation}\:{is} \\ $$$${true}\:{at}\:{term}\:{n}+\mathrm{1} \\ $$

Answered by mathocean1 last updated on 21/Oct/20

Thank you all sirs

$${Thank}\:{you}\:{all}\:{sirs} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com