Question-156979

Question Number 156979 by cortano last updated on 18/Oct/21

Commented by cortano last updated on 18/Oct/21

\underset{k = 1}{\sum^{n}} {(- 1)}^{k} C_{k}^{n} = 0

Commented by cortano last updated on 18/Oct/21

p r o v e t h e e q u a l i t y

Answered by FongXD last updated on 18/Oct/21

Binomial Expansion : {(a + b)}^{n} = \underset{r = 0}{\sum^{n}} C (n, r) a^{n - r} b^{r}

set a = 1 and b = - 1

we get : {(1 - 1)}^{n} = \underset{r = 0}{\sum^{n}} C (n, r) {(- 1)}^{r}

\Leftrightarrow 0 = C (n, 0) + C (n, 1) {(- 1)}^{1} + C (n, 2) {(- 1)}^{2} + C (n, 3) {(- 1)}^{3} + \dots + C (n, n) {(- 1)}^{n}

therefore, 1 - C (n, 1) + C (n, 2) - C (n, 3) + \dots + {(- 1)}^{n} C (n, n) = 0

Answered by puissant last updated on 18/Oct/21

f (x) = {(1 + x)}^{n} = \underset{k = 0}{\sum^{n}} C_{k}^{n} x^{k} 1^{n - k}

\Rightarrow f (x) = {(1 + x)}^{n} = \underset{k = 0}{\sum^{n}} C_{k}^{n} x^{k}; x = - 1

\Rightarrow f (- 1) = {(1 - 1)}^{n} = \underset{k = 0}{\sum^{n}} {(- 1)}^{k} C_{k}^{n}

\Rightarrow \underset{k = 0}{\sum^{n}} {(- 1)}^{k} C_{k}^{n} = 0 \dots .

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