prove by mathematical induction, that for all positive integers n, Σ


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Question Number 69762 by Rio Michael last updated on 27/Sep/19

$p r o v e b y m a t h e m a t i c a l i n d u c t i o n, t h a t f o r a l l p o s i t i v e i n t e g e r s n,$ $\underset{r = 1}{\sum^{n}} r (r + 1) = \frac{n}{3} (n + 1) (n + 2)$
Commented by mathmax by abdo last updated on 27/Sep/19

$1 \times 2 + 2 \times 3 + 3 \times 4 + . . . . + n (n + 1) = \frac{n (n + 1) (n + 2)}{3} (★)$ $n = 1 w e g e t 1 \times 2 = \frac{1 \times 2 \times 3}{3} t h e r e l a t i o n i s t r u e$ $l e t s u p p o s e (★) t r u e w e h a v e$ $1 \times 2 + 2 \times 3 + . . . . + (n + 1) (n + 2) = 1 \times 2 + 2 \times 3 + . . . + n (n + 1)$ $+ (n + 1) (n + 2) = \frac{n (n + 1) (n + 2)}{3} + (n + 1) (n + 2)$ $= (n + 1) (n + 2) (\frac{n}{3} + 1) = \frac{(n + 1) (n + 2) (n + 3)}{3}$ $s o t h e r e l s t i o n i s t r u e a t t e r m (n + 1) .$
	Answered by $@ty@m123 last updated on 27/Sep/19
	$P(k+1): Σ_(r=1) ^(k+1) r(r+1)=(k/3)(k+ 1)( k + 2)+( k + 1)( k + 2) =( k + 1)( k + 2){(k/3)+1} =( k + 1)( k + 2)(((k+3)/3)) =((k+1)/3)( k + 2)( k + 3) ⇒P(k+1) is true.$
	$P (k + 1) :$ $\underset{r = 1}{\sum^{k + 1}} r (r + 1) = \frac{k}{3} (k + 1) (k + 2) + (k + 1) (k + 2)$ $= (k + 1) (k + 2) {\frac{k}{3} + 1}$ $= (k + 1) (k + 2) (\frac{k + 3}{3})$ $= \frac{k + 1}{3} (k + 2) (k + 3)$ $\Rightarrow P (k + 1) i s t r u e .$
	Commented by Rio Michael last updated on 27/Sep/19

	$w a i t s i r [i f p (k + 1) i s t r u e d o w e e n d t h e r e ?$
	Commented by $@ty@m123 last updated on 28/Sep/19

	$N o .$ $I h a v e d o n e t h e d i f f i c u l t p a r t o n l y .$ $I a s u m e d y o u c a n d o t h e r e s t .$
	Commented by Rio Michael last updated on 28/Sep/19

	$t h a n k s$
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