Question Number 16374 by gourav~ last updated on 21/Jun/17
$${The}\:{Value}\:{of}\:{the}\:{sum}..\:\underset{{n}=\mathrm{1}} {\overset{\mathrm{13}} {\sum}}\left({i}^{{n}} +{i}^{{n}+\mathrm{1}} \right),\:{Where}\:{i}=\sqrt{−\mathrm{1}\:\:} \\ $$$${is}.. \\ $$$$\left({a}.\right)\:{i} \\ $$$$\left({b}.\right)\:{i}−\mathrm{1} \\ $$$$\left({c}.\right)\:−{i} \\ $$$$\left({d}.\right)\:\mathrm{0} \\ $$
Answered by Tinkutara last updated on 21/Jun/17
$$\left({i}^{\mathrm{1}} \:+\:{i}^{\mathrm{2}} \right)\:+\:\left({i}^{\mathrm{2}} \:+\:{i}^{\mathrm{3}} \right)\:+\:\left({i}^{\mathrm{3}} \:+\:{i}^{\mathrm{4}} \right)\:+\:…\:+ \\ $$$$\left({i}^{\mathrm{12}} \:+\:{i}^{\mathrm{13}} \right)\:+\:\left({i}^{\mathrm{13}} \:+\:{i}^{\mathrm{14}} \right) \\ $$$$=\:\left({i}^{\mathrm{1}} \:+\:{i}^{\mathrm{14}} \right)\:+\:\mathrm{2}\left({i}^{\mathrm{2}} \:+\:{i}^{\mathrm{3}} \:+\:…\:+\:{i}^{\mathrm{13}} \right) \\ $$$$=\:{i}\:+\:\left(−\mathrm{1}\right)\:=\:{i}\:−\:\mathrm{1} \\ $$$$\left(\mathrm{Using}\:\mathrm{the}\:\mathrm{general}\:\mathrm{rule}:\right. \\ $$$$\left.{i}^{{n}} \:+\:{i}^{{n}+\mathrm{1}} \:+\:{i}^{{n}+\mathrm{2}} \:+\:{i}^{{n}+\mathrm{3}} \:=\:\mathrm{0}\right) \\ $$