Menu Close

show-by-recurrence-that-for-n-N-2-6n-5-3-2n-is-divisible-by-11-

Tinku Tara 0 Comments
Pin




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

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *