How-can-you-prove-not-geometrically-the-following-k-0-n-k-n-n-1-2-

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 + \dots + 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, \dots, 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

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) + \dots . + 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 + \dots \dots \dots \dots + (n - 2) + (n - 1) + n

again \underset{k = 0}{\sum^{n}} k = n + (n - 1) + (n - 2) + \dots \dots \dots \dots . + 3 + 2 + 1

adding 2 \underset{k = 0}{\sum^{n}} = (n + 1) + (n + 1) + \dots . 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 + \dots + 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) + \dots + 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 + \dots (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 .