Show that 1n^3 + 2n + 3n^2 is divisible by 2 and 3 for all positive integers n.


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Question Number 67501 by TawaTawa last updated on 28/Aug/19

$Show that {1 n}^{3} + 2 n + {3 n}^{2} is divisible by 2 and 3 for all positive integers n .$
Commented by Prithwish sen last updated on 28/Aug/19

$Another approch$ $We know that$ ${(1 + x)}^{n} = 1 + nx + \frac{n (n - 1)}{2!} x^{2} + \frac{n (n - 1) (n - 2)}{3!} x^{3 +}$ $. . . . . . . . + \frac{n (n - 1) (n - 2) . . . . . . . (n - r + 1)}{r!} x^{r} + . . . . . + x^{n}$ $Now as far n \in N the coefficients of the expression$ $are integers$ $∴ n (n - 1) must be divisible by 2! = 2$ $n (n - 1) (n - 2)^{″}^{″}^{″}^{″} 3! = 6$ $. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .$ $n (n - 1) (n - 2) . . . . (n - r + 1) must be divisible by r!$ $Now by putting n = n + 2 at n (n - 1) (n - 2) we get,$ $n (n + 1) (n + 2) must be divisible by 3! = 6 proved .$
Commented by TawaTawa last updated on 28/Aug/19

$God bless you sir$
	Answered by Rasheed.Sindhi last updated on 28/Aug/19

	$n^{3} + 3 n^{2} + 2 n = n (n + 1) (n + 3)$ $F o r 2 : n \in E \lor n \in O$ $n \in E \Rightarrow 2 ∣ n \Rightarrow 2 ∣ n (n + 1) (n + 2)$ $\Rightarrow 2 ∣ n^{3} + 3 n^{2} + 2 n$ $n \in O \Rightarrow n + 1 \in E \Rightarrow 2 ∣ n + 1 \Rightarrow 2 ∣ n (n + 1) (n + 2)$ $\Rightarrow 2 ∣ n^{3} + 3 n^{2} + 2 n$ $F o r 3 : n = 3 k o r n = 3 k + 1 o r n = 3 k + 2 \forall k \in Z$ $n = 3 k \Rightarrow 3 ∣ n \Rightarrow 3 ∣ n (n + 1) (n + 2)$ $\Rightarrow 3 ∣ n^{3} + 3 n^{2} + 2 n$ $n = 3 k + 1 \Rightarrow n + 2 = 3 k + 1 + 2 = 3 (k + 1)$ $3 ∣ n + 2 \Rightarrow 3 ∣ n (n + 1) (n + 2)$ $\Rightarrow 3 ∣ n^{3} + 3 n^{2} + 2 n$ $n = 3 k + 2 \Rightarrow n + 1 = 3 k + 3 = 3 (k + 1)$ $\Rightarrow 3 ∣ n + 1 \Rightarrow 3 ∣ n (n + 1) (n + 2)$ $\Rightarrow 3 ∣ n^{3} + 3 n^{2} + 2 n$ $H e n c e 2 ∣ n^{3} + 3 n^{2} + 2 n \land 3 ∣ n^{3} + 3 n^{2} + 2 n$
	Commented by TawaTawa last updated on 28/Aug/19

	$God bless you sir$
	Answered by petrochengula last updated on 28/Aug/19

	$n^{3} + 2 n + 3 n^{2} = n (n + 1) (n + 2)$ $l e t t h e s t a t e m e n t p (n) g i v e n a s p (n) : n^{3} + 2 n + 3 n^{2} i s d i v i s i b l e b y 2 a n d 3, \forall n ϵ z^{+}$ $w e o b s e r v e t h a t p (1) i s t r u e, s i n c e 1^{2} + 2 + 3 = 6 i s d i v i s i b l e b y 2 a n d 3$ $a s s u m e t h a t p (n) i s t r u e f o r s o m e i n t e g e r k$ $p (k) : k^{3} + 2 k + 3 k^{2} i s d i v i s i b l e b y 2 a n d 3 s u c h t h a t k^{3} + 2 k + 3 k^{2} = 2 p a n d k^{3} + 2 k + 3 k^{2} = 3 q w h e r e p, q ϵ Z^{+} . N o w t o p r o v e t h a t p (k + 1) : {(k + 1)}^{3} + 2 (k + 1) + 3 {(k + 1)}^{2}$ $1^{s t} c a s e$ $w e h a v e t o s h o w t h a t p (k + 1) i s d i v i s i b l e b y 2$ ${(k + 1)}^{3} + 2 (k + 1) + 3 {(k + 1)}^{2} = k^{3} + 3 k^{2} + 3 k + 2 k + 2 + 3 (k^{2} + 2 k + 1)$ $= k^{3} + 2 k + 3 k^{2} + 2 (k^{2} + 3 k + 2) + (k + 1) (k + 2)$ $b u t k^{3} + 2 k + 3 k^{2} = 2 p \Rightarrow k (k + 1) (k + 2) = 2 p \Rightarrow (k + 1) (k + 2) = \frac{2 p}{k}$ ${(k + 1)}^{3} + 2 (k + 1) + 3 {(k + 1)}^{2} = 2 p + 2 (k^{2} + 3 k + 2) + \frac{2 p}{k}$ $= 2 (p + k^{2} + 3 k + 2 + \frac{p}{k}) = 2 m$ $2^{n d} c a s e$ $W e h a v e t o s h o w t h a t p (k + 1) i s d i v i s i b l e b y 3$ ${(k + 1)}^{3} + 2 (k + 1) + 3 {(k + 1)}^{2} = k^{3} + 2 k + 3 k^{2} + 3 k^{2} + 9 k + 6$ $= 3 q + 3 (k^{2} + 3 k + 2)$ $= 3 (q + k^{2} + 3 k + 2)$ $T h u s p (k + 1) i s t r u e, w h e n e v e r p (k) i s t r u e$ $H e n c e b y t h e p r i n c i p l e o f m a t h e m a t i c a l i n d u c t i o n p (n) i s t r u e f o r a l l p o s i t i v e i n t e g e r s n u m b e r n .$
	Commented by TawaTawa last updated on 28/Aug/19

	$God bless you sir$
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