prove-that-1-r-1-n-r-1-2-n-n-1-2-r-1-n-r-2-1-6-n-n-1-2n-1-3-r-1-n-r-3-1-4-n-2-n-1-2-

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 \dots \dots 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} = 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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