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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

$$provetheequality$$

Answered by FongXD last updated on 18/Oct/21

$$\mathrm{Binomial}\mathrm{Expansion}:{(\mathrm{a}+\mathrm{b})}^{\mathrm{n}}=\underset{\mathrm{r}=0}{\sum ^{\mathrm{n}}}\mathrm{C}(\mathrm{n},\mathrm{r}){\mathrm{a}}^{\mathrm{n}-\mathrm{r}}{\mathrm{b}}^{\mathrm{r}}$$$$\mathrm{set}\mathrm{a}=1\mathrm{and}\mathrm{b}=-1$$$$\mathrm{we}\mathrm{get}:{(1-1)}^{\mathrm{n}}=\underset{\mathrm{r}=0}{\sum ^{\mathrm{n}}}\mathrm{C}(\mathrm{n},\mathrm{r}){(-1)}^{\mathrm{r}}$$$$\iff 0=\mathrm{C}(\mathrm{n},0)+\mathrm{C}(\mathrm{n},1){(-1)}^1+\mathrm{C}(\mathrm{n},2){(-1)}^2+\mathrm{C}(\mathrm{n},3){(-1)}^3+...+\mathrm{C}(\mathrm{n},\mathrm{n}){(-1)}^{\mathrm{n}}$$$$\mathrm{therefore},1-\mathrm{C}(\mathrm{n},1)+\mathrm{C}(\mathrm{n},2)-\mathrm{C}(\mathrm{n},3)+...+{(-1)}^{\mathrm{n}}\mathrm{C}(\mathrm{n},\mathrm{n})=0$$

Answered by puissant last updated on 18/Oct/21

$$f\left(x\right)={(1+x)}^n=\underset{k=0}{\sum ^n}{C}_k^n{x}^k{1}^{n-k}$$$$\Rightarrow f\left(x\right)={(1+x)}^n=\underset{k=0}{\sum ^n}{C}_k^n{x}^k;x=-1$$$$\Rightarrow f(-1)={(1-1)}^n=\underset{k=0}{\sum ^n}{(-1)}^k{C}_k^n$$$$\Rightarrow \underset{k=0}{\sum ^n}{(-1)}^k{C}_k^n=0....$$

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