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Question Number 108476 by pticantor last updated on 17/Aug/20

$${\mathit{S}}_n=\underset{k=1}{\sum ^n}{k}^2{(-1)}^k{\mathit{C}}_{\mathit{n}}^{\mathit{k}}=?$$$$\mathit{p}\mathit{l}\mathit{e}asehelp$$

Answered by mr W last updated on 17/Aug/20

$${(1-x)}^n=\underset{k=0}{\sum ^n}{(-1)}^k{C}_n^k{x}^k$$$$-n{(1-x)}^{n-1}=\underset{k=0}{\sum ^n}k{(-1)}^k{C}_n^k{x}^{k-1}$$$$-n{(1-x)}^{n-1}x=\underset{k=0}{\sum ^n}k{(-1)}^k{C}_n^k{x}^k$$$$n(n-1){(1-x)}^{n-2}x-n{(1-x)}^{n-1}=\underset{k=0}{\sum ^n}{k}^2{(-1)}^k{C}_n^k{x}^{k-1}$$$$x=1:$$$$0=\underset{k=1}{\sum ^n}{k}^2{(-1)}^k{C}_n^k$$

Answered by mathmax by abdo last updated on 17/Aug/20

$$\mathrm{let}\mathrm{f}\left(\mathrm{x}\right)=\sum _{\mathrm{k}=1}^{\mathrm{n}}{(-1)}^{\mathrm{k}}{\mathrm{C}}_{\mathrm{n}}^{\mathrm{k}}{\mathrm{x}}^{\mathrm{k}}\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\sum _{\mathrm{k}=1}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}^{\mathrm{k}}{(-\mathrm{x})}^{\mathrm{k}}={(1-\mathrm{x})}^{\mathrm{n}}$$$$\Rightarrow \sum _{\mathrm{k}=1}^{\mathrm{n}}\mathrm{k}{(-1)}^{\mathrm{k}}{\mathrm{C}}_{\mathrm{n}}^{\mathrm{k}}{\mathrm{x}}^{\mathrm{k}-1}=-\mathrm{n}{(1-\mathrm{x})}^{\mathrm{n}-1}\Rightarrow$$$$\sum _{\mathrm{k}=1}^{\mathrm{n}}\mathrm{k}{(-1)}^{\mathrm{k}}{\mathrm{C}}_{\mathrm{n}}^{\mathrm{k}}{\mathrm{x}}^{\mathrm{k}}=-\mathrm{nx}{(1-\mathrm{x})}^{\mathrm{n}-1}\Rightarrow$$$$\sum _{\mathrm{k}=1}^{\mathrm{n}}{\mathrm{k}}^2{(-1)}^{\mathrm{k}}{\mathrm{C}}_{\mathrm{n}}^{\mathrm{k}}{\mathrm{x}}^{\mathrm{k}-1}=-\mathrm{n}{(1-\mathrm{x})}^{\mathrm{n}-1}\times \mathrm{n}(\mathrm{n}-1)\mathrm{x}{(1-\mathrm{x})}^{\mathrm{n}-2}$$$$\mathrm{x}=1\Rightarrow \sum _{\mathrm{k}=1}^{\mathrm{n}}{\mathrm{k}}^2{(-1)}^{\mathrm{k}}{\mathrm{C}}_{\mathrm{n}}^{\mathrm{k}}=0$$

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