If-n-an-odd-natural-number-then-r-0-n-1-r-n-C-r-equals-

Question Number 55784 by gunawan last updated on 04/Mar/19

If n an odd natural number, then

\underset{r = 0}{\sum^{n}} \frac{{(- 1)}^{r}}{^{n} C_{r}} equals

Answered by tanmay.chaudhury50@gmail.com last updated on 04/Mar/19

S = T_{0} + T_{1} + T_{2} + \dots + T_{n}

= \frac{{(- 1)}^{0}}{\frac{n!}{0! (n - 0)!}} + \frac{{(- 1)}^{1}}{\frac{n!}{1! (n - 1)!}} + \frac{{(- 1)}^{2}}{\frac{n!}{2! (n - 2)!}} + \frac{{(- 1)}^{3}}{\frac{n!}{3! (n - 3)!}} + \dots + \frac{{(- 1)}^{n}}{\frac{n!}{n! (n - n)!}}

= \frac{0! (n - 0)!}{n!} - \frac{1! (n - 1)!}{n!} + \frac{2! (n - 2)!}{n!} - \frac{3! (n - 3)!}{n!} + . . - \frac{n! (n - n)!}{n!}

n o w l o o k T_{0} = \frac{0! (n - 0)!}{n!} = 1

T_{n} = \frac{- n! (n - n)!}{n!} = - 1

T_{0} + T_{n} = 0

T_{1} = \frac{- 1! (n - 1)!}{n!} = \frac{- 1}{n - 1}

T_{n - 1} = \frac{{(- 1)}^{n - 1} \times (n - 1)! (1)!}{n!} = \frac{1}{n - 1}

T_{1} + T_{n - 1} = 0

n o w

l e t n = 2 k + 1 t e r m s \to k = \frac{n - 1}{2}

m i d d l e t e r m i s {(k + 1)}_{t h} t e r m

k + 1 \to (\frac{n - 1}{2} + 1) \to \frac{n + 1}{2}

k + 1 t h t e r m \to \frac{n + 1}{2} t e r m \to \frac{{(- 1)}^{\frac{n + 1}{2}}}{{n c}_{\frac{n + 1}{2}}}

= \frac{(\frac{n + 1}{2})! (n - \frac{n + 1}{2})!}{n!} \to \frac{(\frac{n + 1}{2})! (\frac{n - 1}{2})!}{n!}

n o w o n a d d i t i o n a l l t e r m s a n c e l l e d e a c h o t h e r

a s e x p l a i n e d a b o v e \dots o n l y o n e t e r m r e m a i n

s o S = \frac{(\frac{n + 1}{2})! (\frac{n - 1}{2})!}{n} \leftarrow t h i s i s t h e a n s w e r