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Question Number 136433 by metamorfose last updated on 21/Mar/21

$$\underset{k=0}{\sum ^n}{C}_{2n+1}^{2k+1}{8}^k=...???$$

Answered by Olaf last updated on 22/Mar/21

$$$$$${(2\sqrt{2}+1)}^{2n+1}=\underset{p=0}{\sum ^{2n+1}}{\mathrm{C}}_p^{2n+1}{\left(2\sqrt{2}\right)}^p$$$${(-2\sqrt{2}+1)}^{2n+1}=\underset{p=0}{\sum ^{2n+1}}{(-1)}^p{\mathrm{C}}_p^{2n+1}{\left(2\sqrt{2}\right)}^p$$$${(1+2\sqrt{2})}^{2n+1}-{(1-2\sqrt{2})}^{2n+1}=$$$$=\underset{p=0}{\sum ^{2n+1}}[1-{(-1)}^p]{\mathrm{C}}_p^{2n+1}{\left(2\sqrt{2}\right)}^p$$$$=2\underset{k=0}{\sum ^n}{\mathrm{C}}_{2k+1}^{2n+1}{\left(2\sqrt{2}\right)}^{2k+1}$$$$=4\sqrt{2}\underset{k=0}{\sum ^n}{\mathrm{C}}_{2k+1}^{2n+1}{8}^k$$$$\Rightarrow \underset{k=0}{\sum ^n}{\mathrm{C}}_{2k+1}^{2n+1}{8}^k=\frac{1}{4\sqrt{2}}[{(1+2\sqrt{2})}^{2n+1}-{(1-2\sqrt{2})}^{2n+1}]$$

Commented by metamorfose last updated on 22/Mar/21

$$thnxsir$$

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