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Question Number 73042 by mathmax by abdo last updated on 05/Nov/19

$$provethatfor(n,p)\in N^{{★}^2}\sum _{k=0}^{p}k{C}_n^{p-k}{C}_n^k=n{C}_{2n-1}^{p-1}$$$$concludethevalueof\sum _{k=0}^{n}k{\left(C_n^k\right)}^2$$

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

$$\mathrm{let}\mathrm{p}\left(\mathrm{x}\right)={(\mathrm{x}+1)}^{\mathrm{n}}=\underset{\mathrm{k}=0}{\sum ^{\mathrm{n}}}{\mathrm{C}}_{\mathrm{n}}^{\mathrm{k}}{\mathrm{X}}^{\mathrm{k}}$$$$\mathrm{p}\left(\mathrm{x}\right).{\mathrm{p}}^{\prime }\left(\mathrm{x}\right)=\underset{\mathrm{k}=0}{\sum ^{\mathrm{n}}}\underset{\mathrm{j}=1}{\sum ^{\mathrm{n}}}{\mathrm{C}}_{\mathrm{k}}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{j}}^{\mathrm{n}}{\mathrm{x}}^{\mathrm{k}}.{\mathrm{jx}}^{\mathrm{j}-1}$$$$\mathrm{coeficient}\mathrm{of}\mathrm{power}\mathrm{p}-1\mathrm{is}\mathrm{when}\mathrm{k}+\mathrm{j}=\mathrm{p}$$$$=\underset{\mathrm{k}=0}{\sum ^{\mathrm{n}}}{\mathrm{jC}}_{\mathrm{n}}^{\mathrm{p}-\mathrm{j}}{\mathrm{C}}_{\mathrm{n}}^{\mathrm{j}}{\mathrm{X}}^{\mathrm{p}-1}$$$${\mathrm{p}}^2={(\mathrm{x}+1)}^{2\mathrm{n}}$$$$\text{Prime causes double exponent: use braces to clarify}$$$$\mathrm{coeficent}\mathrm{of}{\mathrm{x}}^{\mathrm{p}-1}\mathrm{is}{\mathrm{nC}}_{2\mathrm{n}-1}^{\mathrm{p}-1}$$$$\Rightarrow \underset{\mathrm{k}=0}{\sum ^{\mathrm{p}}}{\mathrm{kC}}_{\mathrm{n}}^{\mathrm{p}-\mathrm{k}}{\mathrm{C}}_{\mathrm{n}}^{\mathrm{k}}={\mathrm{nC}}_{2\mathrm{n}-1}^{\mathrm{p}-1}$$$$\mathrm{if}\mathrm{p}=\mathrm{n}$$$$\Rightarrow \mathrm{\Sigma }\mathrm{k}{\left({\mathrm{C}}_{\mathrm{n}}^{\mathrm{k}}\right)}^2={\mathrm{nC}}_{2\mathrm{n}-1}^{\mathrm{n}-1}$$

Commented by mr W last updated on 06/Nov/19

$$ilearntanewmethod.thankssir!$$

Commented by mind is power last updated on 06/Nov/19

$${\mathrm{y}}^{\prime }\mathrm{re}\mathrm{welcom}$$

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