Show that for n∈ N, A_n =n^2 (n^2 −1) is divisible by 12


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Question Number 142310 by mathocean1 last updated on 29/May/21

$Show that for n \in N, A_{n} = n^{2} (n^{2} - 1)$ $is divisible by 12$
	Answered by MJS_new last updated on 29/May/21

	$A_{6 k} = 36 k^{2} (6 k - 1) (6 k + 1)$ $A_{6 k + 1} = 12 k (3 k + 1) {(6 k + 1)}^{2}$ $A_{6 k + 2} = 12 (12 k + 1) {(3 k + 1)}^{2} (6 k + 1)$ $A_{6 k + 3} = 36 {(2 k + 1)}^{2} (3 k + 1) (3 k + 2)$ $A_{6 k + 4} = 12 (2 k + 1) {(3 k + 2)}^{2} (6 k + 5)$ $A_{6 k + 5} = 12 (k + 1) (3 k + 2) {(6 k + 5)}^{2}$
	Answered by JDamian last updated on 29/May/21

	$n^{2} (n^{2} - 1) = (n - 1) n^{2} (n + 1)$ $(n - 1) n (n + 1) i s d i v i s i b l e b y 3$ $a n d$ $c o n t a i n s t h e p r o d u c t o f e i t h e r t w o e v e n$ $n u m b e r s (n - 1 a n d n + 1) o r o n l y o n e$ $e v e n n u m b e r n, w h i c h a p p e a r s a s n^{2} i n A_{n} .$ $I n a n y c a s e, A_{n} i s a l s o d i v i s i b l e b y 4 .$
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