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

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

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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