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Question Number 163263 by Ar Brandon last updated on 05/Jan/22

Answered by qaz last updated on 05/Jan/22

$$\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}}}{(-1)}^{\mathrm{k}}\mathrm{k}\left(\begin{array}{c}\mathrm{n}\\ \mathrm{k}\end{array}\right)=\frac{\partial }{\partial \mathrm{x}}{\mid }_{\mathrm{x}=1}(\underset{\mathrm{k}=0}{\sum ^{\mathrm{n}}}\left(\begin{array}{c}\mathrm{n}\\ \mathrm{k}\end{array}\right){(-\mathrm{x})}^{\mathrm{k}}-1)=\frac{\partial }{\partial \mathrm{x}}{\mid }_{\mathrm{x}=1}{(1-\mathrm{x})}^{\mathrm{n}}=-\mathrm{n}{(1-\mathrm{x})}^{\mathrm{n}-1}{\mid }_{\mathrm{x}=1}=0$$$$\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}}}\mathrm{k}\left(\begin{array}{c}\mathrm{n}\\ \mathrm{k}\end{array}\right)=\frac{\partial }{\partial \mathrm{x}}{\mid }_{\mathrm{x}=1}(\underset{\mathrm{k}=0}{\sum ^{\mathrm{n}}}\left(\begin{array}{c}\mathrm{n}\\ \mathrm{k}\end{array}\right){\mathrm{x}}^{\mathrm{k}}-1)=\frac{\partial }{\partial \mathrm{x}}{\mid }_{\mathrm{x}=1}{(1+\mathrm{x})}^{\mathrm{n}}=\mathrm{n}{(1+\mathrm{x})}^{\mathrm{n}-1}{\mid }_{\mathrm{x}=1}={\mathrm{n}2}^{\mathrm{n}-1}$$$$\underset{\mathrm{By}{\mathrm{Egyolychev}}^{\prime }\mathrm{s}\mathrm{method}}{\underbrace{\underset{\mathrm{k}=0}{\sum ^{\mathrm{n}}}{\left(\begin{array}{c}\mathrm{n}\\ \mathrm{k}\end{array}\right)}^2=\underset{\mathrm{k}=0}{\sum ^{\mathrm{n}}}\left(\begin{array}{c}\mathrm{n}\\ \mathrm{k}\end{array}\right)\left(\begin{array}{c}\mathrm{n}\\ \mathrm{n}-\mathrm{k}\end{array}\right)=\left[{\mathrm{z}}^{\mathrm{n}}\right]{(1+\mathrm{z})}^{\mathrm{n}}\underset{\mathrm{k}=0}{\sum ^{\mathrm{n}}}\left(\begin{array}{c}\mathrm{n}\\ \mathrm{k}\end{array}\right){\mathrm{z}}^{\mathrm{k}}=\left[{\mathrm{z}}^{\mathrm{n}}\right]{(1+\mathrm{z})}^{2\mathrm{n}}=\left(\begin{array}{c}2\mathrm{n}\\ \mathrm{n}\end{array}\right)}}$$

Commented by Ar Brandon last updated on 06/Jan/22

thanks

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