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Question Number 159259 by pticantor last updated on 14/Nov/21

$$\mathit{m}\mathit{o}\mathit{n}\mathit{t}\mathit{r}\mathit{e}\mathit{q}\mathit{u}\mathit{e}$$$$\underset{\mathit{k}=0}{\sum ^{\mathit{n}}}{\mathit{C}}_{2\mathit{n}}^{2\mathit{k}}=2^{2\mathit{n}-1}$$

Answered by mr W last updated on 14/Nov/21

$${(1+x)}^{2n}=\underset{k=0}{\sum ^n}{C}_{2k}^{2n}{x}^{2k}+\underset{k=0}{\sum ^{n-1}}{C}_{2k+1}^{2n}{x}^{2k+1}..\left(i\right)$$$${(1-x)}^{2n}=\underset{k=0}{\sum ^n}{C}_{2k}^{2n}{x}^{2k}-\underset{k=0}{\sum ^{n-1}}{C}_{2k+1}^{2n}{x}^{2k+1}..\left(ii\right)$$$$$$$$\underset{k=0}{\sum ^n}{C}_{2k}^{2n}{x}^{2k}=\frac{{(1+x)}^{2n}+{(1-x)}^{2n}}{2}$$$$letx=1$$$$\underset{k=0}{\sum ^n}{C}_{2k}^{2n}=\frac{{(1+1)}^{2n}+{(1-1)}^{2n}}{2}=\frac{2^{2n}}{2}=2^{2n-1}$$

Commented by pticantor last updated on 15/Nov/21

$$\mathit{m}\mathit{e}\mathit{r}\mathit{c}\mathit{i}\mathit{b}\mathit{e}\mathit{a}\mathit{u}\mathit{c}\mathit{o}\mathit{u}\mathit{p}$$

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