k-1-2n-1-k-A-B-1-C-1-D-0-

Question Number 76813 by Rio Michael last updated on 30/Dec/19

\underset{k = 1}{\sum^{2 n}} {(- 1)}^{k} =

A . \infty

B . 1

C . - 1

D . 0

Commented by JDamian last updated on 30/Dec/19

D

Commented by Rio Michael last updated on 30/Dec/19

why sir ?

Commented by JDamian last updated on 30/Dec/19

S = - 1, 0, - 1, 0, \cdot \cdot \cdot

S_{n} = {\begin{cases} - 1 w h e n n i s o d d \\ 0 w h e n n i s e v e n \end{cases}

A s 2 n i s a l w a y s e v e n, t h e n S_{n} = 0

Commented by Rio Michael last updated on 30/Dec/19

sir k = 1 \Rightarrow S = - 1

k = 2 \Rightarrow S = 1

S = - 1, 1, - 1, 1 \dots ?

Commented by JDamian last updated on 30/Dec/19

Y o u a r e c o n f u s e d - I d e p i c t e d S_{n} a s t h e

s u m o f t h e s q u e n c e a_{n} = {(- 1)}^{n} u p t o n . B u t

y o u a r e u s i n g S_{n} a s t h e s e q u e n c e a_{n} i t s e l f .

T h i s p r o b l e m i s a b o u t t h e s u m o f t h e

{(- 1)}^{n} s e q u e n c e .

Commented by Rio Michael last updated on 30/Dec/19

thanks sir .

Commented by abdomathmax last updated on 30/Dec/19

S_{n} = \sum_{k = 1}^{2 n} {(- 1)}^{k} = \sum_{k = 0}^{2 n} {(- 1)}^{k} - 1

= \frac{1 - {(- 1)}^{2 n + 1}}{1 - (- 1)} - 1 = \frac{2}{2} - 1 = 0

s o t h e c o r r e c t a n s w e r i s D

Answered by ~blr237~ last updated on 30/Dec/19

S_{n} = 1 + x + \dots + x^{n} = \frac{1 - x^{n + 1}}{1 - x}

\underset{k = 1}{\sum^{2 n}} {(- 1)}^{k} = (- 1) + \underset{k = 0}{\sum^{2 n}} {(- 1)}^{k} = - 1 + \frac{1 - {(- 1)}^{2 n + 1}}{1 - (- 1)} = - 1 + 1 = 0

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