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


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

$s h o w b y r e c u r r e n c e t h a t :$ $f o r n \in N^{*}, 2^{6 n - 5} + 3^{2 n} i s d i v i s i b l e b y$ $11 .$
	Answered by Dwaipayan Shikari last updated on 20/Oct/20

	$2^{6 n - 5} + 3^{2 n} = 11 d = Ψ (n)$ $2^{6 n + 1} + 9^{n + 1} = Ψ (n + 1) (R e p l a c i n g n a s n + 1)$ $= 2^{6 n + 1} + 2 . 2^{6 n - 5} + 9^{n + 1} + 2 . 9^{n} - 2 (2^{6 n - 5} + 9^{n})$ $= 2^{6 n - 5} (2^{6} + 2) + 9^{n} (9 + 2) - 22 d (A s Ψ (n) = 2^{6 n - 5} + 3^{2 n} = 11 d)$ $= 66 . 2^{6 n - 5} + 11 . 9^{n} - 22 d$ $W h i c h i s d i v i s i b l e b y 11$ $B o t h Ψ (n) a n d Ψ (n + 1) i s d i v i s i b l e b y 11$ $S o$ $G e n e r a l l y Ψ (n) = 2^{6 n - 5} + 3^{2 n} i s d i v i s b l e b y 11$
	Answered by Bird last updated on 21/Oct/20
	$let u_n =2^(6n−5 ) +3^(2n) n=1 ⇒u_1 =2+3^2 =11 this number is divisible by 11 let suppose u_n divisible by 11 we have u_(n+1) =2^(6n+6−5) +3^(2n+2) =2^6 (u_n −3^(2n) )+3^(2n) ×9 =2^6 u_n −2^6 3^(2n) +3^(2n) ×9 =2^6 u_n +3^(2n) (9−2^6 ) =2^6 (11k)+55.3^(2n) =11{2^6 k+5.3^(2n) } ⇒u_(n+1) is divisible by 11 the telation is true at term n+1$
	$l e t u_{n} = 2^{6 n - 5} + 3^{2 n}$ $n = 1 \Rightarrow u_{1} = 2 + 3^{2} = 11 t h i s n u m b e r$ $i s d i v i s i b l e b y 11$ $l e t s u p p o s e u_{n} d i v i s i b l e b y 11$ $w e h a v e u_{n + 1} = 2^{6 n + 6 - 5} + 3^{2 n + 2}$ $= 2^{6} (u_{n} - 3^{2 n}) + 3^{2 n} \times 9$ $= 2^{6} u_{n} - 2^{6} 3^{2 n} + 3^{2 n} \times 9$ $= 2^{6} u_{n} + 3^{2 n} (9 - 2^{6})$ $= 2^{6} (11 k) + 55 . 3^{2 n}$ $= 11 {2^{6} k + 5 . 3^{2 n}} \Rightarrow u_{n + 1} i s$ $d i v i s i b l e b y 11 t h e t e l a t i o n i s$ $t r u e a t t e r m n + 1$
	Answered by mathocean1 last updated on 21/Oct/20

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