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Question Number 16374 by gourav~ last updated on 21/Jun/17

The Value of the sum.. Σ_(n=1) ^(13) (i^n +i^(n+1) ), Where i=(√(−1  ))  is..  (a.) i  (b.) i−1  (c.) −i  (d.) 0

$${The}\:{Value}\:{of}\:{the}\:{sum}..\:\underset{{n}=\mathrm{1}} {\overset{\mathrm{13}} {\sum}}\left({i}^{{n}} +{i}^{{n}+\mathrm{1}} \right),\:{Where}\:{i}=\sqrt{−\mathrm{1}\:\:} \\ $$$${is}.. \\ $$$$\left({a}.\right)\:{i} \\ $$$$\left({b}.\right)\:{i}−\mathrm{1} \\ $$$$\left({c}.\right)\:−{i} \\ $$$$\left({d}.\right)\:\mathrm{0} \\ $$

Answered by Tinkutara last updated on 21/Jun/17

(i^1  + i^2 ) + (i^2  + i^3 ) + (i^3  + i^4 ) + ... +  (i^(12)  + i^(13) ) + (i^(13)  + i^(14) )  = (i^1  + i^(14) ) + 2(i^2  + i^3  + ... + i^(13) )  = i + (−1) = i − 1  (Using the general rule:  i^n  + i^(n+1)  + i^(n+2)  + i^(n+3)  = 0)

$$\left({i}^{\mathrm{1}} \:+\:{i}^{\mathrm{2}} \right)\:+\:\left({i}^{\mathrm{2}} \:+\:{i}^{\mathrm{3}} \right)\:+\:\left({i}^{\mathrm{3}} \:+\:{i}^{\mathrm{4}} \right)\:+\:...\:+ \\ $$$$\left({i}^{\mathrm{12}} \:+\:{i}^{\mathrm{13}} \right)\:+\:\left({i}^{\mathrm{13}} \:+\:{i}^{\mathrm{14}} \right) \\ $$$$=\:\left({i}^{\mathrm{1}} \:+\:{i}^{\mathrm{14}} \right)\:+\:\mathrm{2}\left({i}^{\mathrm{2}} \:+\:{i}^{\mathrm{3}} \:+\:...\:+\:{i}^{\mathrm{13}} \right) \\ $$$$=\:{i}\:+\:\left(−\mathrm{1}\right)\:=\:{i}\:−\:\mathrm{1} \\ $$$$\left(\mathrm{Using}\:\mathrm{the}\:\mathrm{general}\:\mathrm{rule}:\right. \\ $$$$\left.{i}^{{n}} \:+\:{i}^{{n}+\mathrm{1}} \:+\:{i}^{{n}+\mathrm{2}} \:+\:{i}^{{n}+\mathrm{3}} \:=\:\mathrm{0}\right) \\ $$

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