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What is the value of $\; \large\frac{i^{4n+1} - i^{4n-1}}{2}\;$ ?

$(a)\;i\qquad(b)\;-i\qquad(c)\;2i\qquad(d)\;4i$

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Answer : $\;i$
Explanation :
$\; \large\frac{i^{4n+1} - i^{4n-1}}{2}\; = \large\frac{i^{4n} . i - i^{4n} . i^{-1}}{2}$
$= \large\frac{i - \large\frac{1}{i}}{2} = \large\frac{i^{2}-1}{2i}$
$= \large\frac{-2}{2i}$
$= \large\frac{-1}{i} = i \;.$
answered Apr 18, 2014 by yamini.v
 

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