prove by mathematical induction that n(n+1)(n+2) is an integer multiple of 6


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Question Number 117903 by aurpeyz last updated on 14/Oct/20

$p r o v e b y m a t h e m a t i c a l i n d u c t i o n$ $t h a t n (n + 1) (n + 2) i s a n i n t e g e r$ $m u l t i p l e o f 6$
	Answered by mindispower last updated on 14/Oct/20

	$(n + 1) (n + 2) (n + 3)$ $= n (n + 1) (n + 2) + 3 (n + 1) (n + 2)$ $= n (n + 1) (n + 2) + 3 n^{2} + 9 n + 6$ $= n (n + 1) (n + 2) + 3 n (n + 1) + 6 (n + 1)$ $p_{m} = m (m + 1) (m + 2)$ $2 ∣ n (n + 1) ‘ ‘ n = 2 k t r u, n = 2 k + 1 \Rightarrow n + 1 = 2 (k + 1) {t r u e}^{″}$ $f o r n = 0 w e h v e 6 ∣ 0 t r u e$ $s u p o s e \forall n 6 ∣ P_{n}; 6 ∣ P_{n + 1}$ $P_{n + 1} = (n + 1) (n + 2) (n + 3) = n (n + 1) (n + 2) + 3 n (n + 1) + 6$ $p_{n} = n (n + 1) (n + 2) = 6 k b y h y p o t h e s$ $n (n + 1) = 2 a$ $\Rightarrow p_{n + 1} = 6 (k + a + 1) \Rightarrow 6 ∣ p_{n + 1}$
	Answered by 1549442205PVT last updated on 14/Oct/20

	$Prove n (n + 1) (n + 2) is multiply of 6$ $\forall n \in N, n ⩾ 1 by mathematical inuction$ $i) For n = 1 \Rightarrow A_{1} = 1.2.3 = 6 ⋮ 6 \Rightarrow State is true$ $ii) Suppose State was true \forall n ⩽ k$ $that means A_{k} = k (k + 1) (k + 2) ⋮ 6$ $iii) We will prove that A_{k + 1} ⋮ 6$ $Indeed, A_{k + 1} = (k + 1) (k + 2) (k + 3)$ $= k (k + 1) (k + 2) + 3 (k + 1) (k + 2)$ $= A_{k} + 3 (k^{2} + 3 k + 2) = A_{k} + 3 k (k + 1)$ $+ 6 (k + 1) ⋮ 6 Since A_{k} ⋮ 6, 6 (k + 1) ⋮ 6$ $3 k (k + 1) ⋮ 6 (by in two sequence numbers$ $always have one even number)$ $Thus, State is also for n = k + 1, so$ $by mathimatical induction principle$ $State is true \forall n \in N^{*} (q . e . d)$
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