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Question Number 30519 by abdo imad last updated on 22/Feb/18
$${let}\:{j}={e}^{{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} \:\:\:\:{find}\:{the}\:{value}\:{of}\:\sum_{{k}=\mathrm{0}} ^{{n}} {C}_{{n}} ^{{k}} \left(\mathrm{1}+{j}\right)^{{k}} {j}^{\mathrm{2}{n}−\mathrm{2}{k}} \:. \\ $$
Answered by sma3l2996 last updated on 23/Feb/18
$$\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{C}_{{n}} ^{{k}} \left(\mathrm{1}+{j}\right)^{{k}} {j}^{\mathrm{2}{n}−\mathrm{2}{k}} =\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{C}_{{n}} ^{{k}} \left(\mathrm{1}+{j}\right)^{{k}} {j}^{\mathrm{2}\left({n}−{k}\right)} \\ $$$$=\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{C}_{{n}} ^{{k}} \left(\mathrm{1}+{j}\right)^{{k}} \left({j}^{\mathrm{2}} \right)^{{n}−{k}} =\left(\mathrm{1}+{j}+{j}^{\mathrm{2}} \right)^{{n}} =\mathrm{0}\:\:{because}\:\left({j}^{\mathrm{2}} +{j}+\mathrm{1}=\mathrm{0}\right) \\ $$