How can you prove (not geometrically) the following? Σ


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Question Number 56229 by Hassen_Timol last updated on 12/Mar/19

$How can you prove (not geometrically)$ $the following ?$ $\underset{k = 0}{\sum^{n}} k = \frac{n (n + 1)}{2}$
Commented by Kunal12588 last updated on 12/Mar/19

$C a n I u s e A r i t m e t i c P r o g r e s s i o n A P s i r .$
Commented by Hassen_Timol last updated on 12/Mar/19

$y e s S i r p l e a s e$
Commented by Kunal12588 last updated on 12/Mar/19

$\underset{k = 0}{\sum^{n}} k = 0 + 1 + 2 + 3 + 4 + . . . + n$ $l e t u s w r i t e t h e t e r m s$ $0, 1, 2, 3, 4, 5, 6, 7, . . ., n$ $a b o v e s e q u e n c e i s i n A P$ $w h e r e a = 0, d = 1$ $S_{r} = \frac{r}{2} [2 a + (r - 1) d]$ $∴ S_{n + 1} = \frac{(n + 1)}{2} [2 \times 0 + ((n + 1) - 1) (1)]$ $\Rightarrow S_{n + 1} = \frac{n + 1}{2} (0 + n \times 1) = \frac{n (n + 1)}{2}$
Commented by maxmathsup by imad last updated on 13/Mar/19
$solution by polynome method let p(x)=((x(x−1))/2) we have for all x from R p(x+1)−p(x)=(((x+1)x)/2) −((x(x−1))/2) =((x^2 +x −x^2 +x)/2) =x ⇒ ∀ k∈[[1,n]] p(k+1)−p(k) =k ⇒Σ_(k=1) ^n {p(k+1)−p(k)}=Σ_(k=1) ^n k ⇒ Σ_(k=1) ^n k =p(2)−p(1)+p(3)−p(2)+....+p(n+1)−p(n)=p(n+1)−p(1) =((n(n+1))/2) −0 ⇒ Σ_(k=1) ^n k =((n(n+1))/2) .$
$s o l u t i o n b y p o l y n o m e m e t h o d l e t p (x) = \frac{x (x - 1)}{2} w e h a v e f o r a l l x f r o m R$ $p (x + 1) - p (x) = \frac{(x + 1) x}{2} - \frac{x (x - 1)}{2} = \frac{x^{2} + x - x^{2} + x}{2} = x \Rightarrow$ $\forall k \in [[1, n]] p (k + 1) - p (k) = k \Rightarrow \sum_{k = 1}^{n} {p (k + 1) - p (k)} = \sum_{k = 1}^{n} k \Rightarrow$ $\sum_{k = 1}^{n} k = p (2) - p (1) + p (3) - p (2) + . . . . + p (n + 1) - p (n) = p (n + 1) - p (1)$ $= \frac{n (n + 1)}{2} - 0 \Rightarrow \sum_{k = 1}^{n} k = \frac{n (n + 1)}{2} .$
	Answered by Prithwish sen last updated on 12/Mar/19

	$\underset{k = 0}{\sum^{n}} k = 1 + 2 + 3 + . . . . . . . . . . . . + (n - 2) + (n - 1) + n$ $again \underset{k = 0}{\sum^{n}} k = n + (n - 1) + (n - 2) + . . . . . . . . . . . . . + 3 + 2 + 1$ $adding 2 \underset{k = 0}{\sum^{n}} = (n + 1) + (n + 1) + . . . . upto n times$ $= n (n + 1)$ $∴ \underset{k = 0}{\sum^{n}} = \frac{n (n + 1)}{2} proved$
	Commented by Hassen_Timol last updated on 12/Mar/19

	$Thank you Sir, God bless you$
	Answered by Kunal12588 last updated on 12/Mar/19

	$S_{n + 1} = \underset{k = 0}{\sum^{n}} k = 0 + 1 + 2 + 3 + 4 + . . . + n (1)$ $T h e r e a r e (n + 1) t e r m s i n a b o v e s e r i e s .$ $R e v e r s i n g S_{n} ∵ a d d i t i o n f o l l o w s c o m m u t i v e$ $p r o p e r t y i t d o e s n o t e f f e c t t h e S_{n}$ $S_{n + 1} = n + (n - 1) + (n - 2) + . . . + 0 (2)$ $a d d i n g c o r r e s p o n d i n g t e r m s o f (1) a n d (2)$ $2 S_{n + 1} = (0 + n) + [1 + (n - 1)] + [2 + (n - 2)] + . . + (n + 0)$ $t h e r e a r e (n + 1) b r a c k e t s$ $a s y o u c a n s e e e v e r y b r a c k e t w i l l g i v e u s^{'} n^{'}$ $2 S_{n + 1} = n + n + n + n + . . . (n + 1) t e r m s$ $2 S_{n + 1} = n (n + 1)$ $S_{n + 1} = \frac{n (n + 1)}{2}$
	Commented by Hassen_Timol last updated on 12/Mar/19

	$Thank you Sir, God bless you$
	Commented by Prithwish sen last updated on 12/Mar/19

	$this is correct$
	Commented by Kunal12588 last updated on 12/Mar/19

	$M a y G o d b l e s s y o u t o o .$ $W e l c o m e .$
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