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Question Number 73041 by mathmax by abdo last updated on 05/Nov/19

$$provethat\mathrm{\forall }n\in N\sum _{k=0}^{2n}{(-1)}^k{\left(C_{2n}^k\right)}^2={(-1)}^n{C}_{2n}^n$$

Commented by mathmax by abdo last updated on 06/Nov/19

$$letp\left(x\right)={(1+x)}^{2n}andq\left(x\right)={(1-x)}^{2n}\Rightarrow$$$$p\left(x\right).q\left(x\right)=\left(\sum _{k=0}^{2n}{C}_{2n}^k{x}^k\right)\left(\sum _{k=0}^{2n}{C}_{2n}^k{(-1)}^k{x}^k\right)$$$$=\sum _{k=0}^{2n}{C}_k{x}^{k}with{C}_k=\sum _{i+j=k}{a}_i{b}_j=\sum _{i=0}^k{a}_i{b}_{k-i}$$$$=\sum _{i=0}^k{C}_{2n}^i{(-1)}^{k-i}{C}_{2n}^{k-i}\Rightarrow thecoefficientof{x}^{2n}is$$$$C_{2n}=\sum _{i=0}^{2n}{C}_{2n}^i{(-1)}^{2n-i}{C}_{2n}^{2n-i}=\sum _{i=0}^{2n}{(-1)}^i{\left(C_{2n}^i\right)}^{2}alsowehave$$$$p\left(x\right)q\left(x\right)={(1-x^2)}^{2n}=\sum _{k=0}^{2n}{(-1)}^k{C}_{2n}^k{x}^{2k}\Rightarrow thecoefficientof{x}^{2n}is$$$$is{(-1)}^n{C}_{2n}^n\Rightarrow \sum _{i=0}^{2n}{(-1)}^i{\left(C_{2n}^i\right)}^2={(-1)}^n{C}_{2n}^n$$

Answered by mind is power last updated on 05/Nov/19

$$\mathrm{let}\mathrm{p}\left(\mathrm{x}\right)={(1+\mathrm{x})}^{2\mathrm{n}}$$$$\mathrm{Q}\left(\mathrm{x}\right)={(1-\mathrm{x})}^{2\mathrm{n}}$$$$\mathrm{p}\left(\mathrm{x}\right)=\underset{\mathrm{k}=0}{\sum ^{\mathrm{n}}}{\mathrm{C}}_{2\mathrm{n}}^{\mathrm{k}}{\mathrm{X}}^{\mathrm{k}}$$$$\mathrm{Q}\left(\mathrm{x}\right)=\underset{\mathrm{k}=0}{\sum ^{\mathrm{n}}}{\mathrm{C}}_{\mathrm{n}}^{\mathrm{k}}{(-\mathrm{x})}^{\mathrm{k}}$$$$\mathrm{lets}\mathrm{find}\mathrm{coeficient}\mathrm{of}{\mathrm{X}}^{2\mathrm{n}}\mathrm{in}\mathrm{p}.\mathrm{Q}$$$$\mathrm{P},\mathrm{Q}=\underset{\mathrm{k}=0}{\sum ^{2\mathrm{n}}}{\mathrm{C}}_{2\mathrm{n}}^{\mathrm{k}}{(-1)}^{\mathrm{k}}{\mathrm{x}}^{\mathrm{k}}.\underset{\mathrm{j}=0}{\sum ^{2\mathrm{n}}}{\mathrm{C}}_{2\mathrm{n}}^{\mathrm{j}}{\mathrm{x}}^{\mathrm{j}}$$$$\Rightarrow \mathrm{k}+\mathrm{j}=2\mathrm{n}$$$$=\underset{\mathrm{k}=0}{\sum ^{2\mathrm{n}}}{(-1)}^{\mathrm{k}}{\mathrm{C}}_{2\mathrm{n}}^{\mathrm{k}}.{\mathrm{C}}_{2\mathrm{n}}^{2\mathrm{n}-\mathrm{k}}=\underset{\mathrm{k}=0}{\sum ^{2\mathrm{n}}}{(-1)}^{\mathrm{k}}{\left({\mathrm{C}}_{2\mathrm{n}}^{\mathrm{k}}\right)}^2$$$$\mathrm{p}.\mathrm{Q}={(1-{\mathrm{x}}^2)}^{2\mathrm{n}}$$$$\mathrm{coeficient}\mathrm{of}{\mathrm{X}}^{2\mathrm{n}}{\mathrm{isC}}_{2\mathrm{n}}^{\mathrm{n}}.{(-1)}^{\mathrm{n}}$$$$\Rightarrow \underset{\mathrm{k}=0}{\sum ^{2\mathrm{n}}}{(-1)}^{\mathrm{k}}{\left({\mathrm{C}}_{2\mathrm{n}}^{\mathrm{k}}\right)}^2={(-1)}^{\mathrm{n}}.{\mathrm{C}}_{2\mathrm{n}}^{\mathrm{n}}$$

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