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Find the value of $ \sum_{n=1}^{200}i^n$

$\begin{array}{1 1}(A) \;0\\(B)\;1\\(C)\;i\\(D)\;200\end{array}$

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1 Answer

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  • $i^4=i^8=i^{12}=.........i^{4n}=1$
  • $1+i+i^2+i^3=0$
$\sum_{n=1}^{200}i^n=i+i^2+i^3+.............i^{200}$
$=(i+i^2+i^3+i^4)+i^4(i+i^2+i^3+i^4)+i^8(i+i^2+i^3+i^4)+......$
$=(i+i^2+i^3+1)+1(i+i^2+i^3+1)+...........$
=0
Hence answer is (A)
answered Jul 4, 2013 by rvidyagovindarajan_1
edited Jul 28, 2014 by sharmaaparna1
 

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