prove that 1. Σ_(r=1) ^n r = (1/2)n(n+1) 2. Σ_(r=1) ^n r^2 = (1/6)n(n+1)(2n + 1) 3. Σ


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Question Number 76368 by Rio Michael last updated on 26/Dec/19

$p r o v e t h a t$ $1 . \underset{r = 1}{\sum^{n}} r = \frac{1}{2} n (n + 1)$ $2 . \underset{r = 1}{\sum^{n}} r^{2} = \frac{1}{6} n (n + 1) (2 n + 1)$ $3 . \underset{r = 1}{\sum^{n}} r^{3} = \frac{1}{4} n^{2} {(n + 1)}^{2}$
Commented by mr W last updated on 27/Dec/19

Commented by Rio Michael last updated on 27/Dec/19

$t h a n k s s i r$
Commented by john santu last updated on 27/Dec/19

$h a h a h a h a . . . . . . c o m p l e t e s i r$
Commented by john santu last updated on 27/Dec/19

$b u t s i r t h e e q u a t i o n h o w t o p r o v e$ $n o t l i s t t h e f o r m u l a . h a h a h g r e a t s i r$
Commented by mr W last updated on 27/Dec/19

$t h e r e a r e m a n y w a y s t o c a l c u l a t e$ $(t o p r o v e) . i f y o u a l r e a d y k n o w S_{k},$ $t h e n y o u c a n a l s o g e t S_{k + 1} .$
	Answered by john santu last updated on 27/Dec/19

	$(1) {(r + 1)}^{2} - r^{2} = 2 r + 1$ $\underset{r = 1}{\sum^{n}} {(r + 1)}^{2} - r^{2} = \underset{r = 1}{\sum^{n}} (2 r + 1)$ ${(n + 1)}^{2} - 1 = 2 \underset{r = 1}{\sum^{n}} (r) + n$ $n^{2} + n = 2 \underset{r = 1}{\sum^{n}} r$ $\underset{r = 1}{\sum^{n}} r = \frac{n^{2} + n}{2} ◼$
	Answered by john santu last updated on 27/Dec/19
	$(2)(r+1)^3 −r^3 =3r^2 +3r+1 Σ_(r=1) ^n {(r+1)^3 −r^3 }=Σ_(r=1) ^n (3r^2 +3r+1) (n+1)^3 −n^3 =3Σ_(r=1) ^n r^2 + ((3n(n+1))/2)+n Σ_(r=1) ^n r^2 =((n(n+1)(2n+1))/6) ■$
	$(2) {(r + 1)}^{3} - r^{3} = 3 r^{2} + 3 r + 1$ $\underset{r = 1}{\sum^{n}} {{(r + 1)}^{3} - r^{3}} = \underset{r = 1}{\sum^{n}} (3 r^{2} + 3 r + 1)$ ${(n + 1)}^{3} - n^{3} = 3 \underset{r = 1}{\sum^{n}} r^{2} + \frac{3 n (n + 1)}{2} + n$ $\underset{r = 1}{\sum^{n}} r^{2} = \frac{n (n + 1) (2 n + 1)}{6} ◼$
	Answered by benjo last updated on 27/Dec/19

	$prove (3) use {(r + 1)}^{4} - r^{4} = {4 r}^{3} + {6 r}^{2} + 4 r + 1$
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