A) \[i\]
B) \[i-1\]
C) \[-i\]
D) 0
Correct Answer: B
Solution :
\[\sum\limits_{n=1}^{13}{({{i}^{n}}+{{i}^{n+1}})}\]\[=(i+{{i}^{2}}+{{i}^{3}}+....+{{i}^{13}})+({{i}^{2}}+{{i}^{3}}+....+{{i}^{14}})\] \[=\frac{i(1-{{i}^{13}})}{1-i}+\frac{{{i}^{2}}(1-{{i}^{13}})}{1-i}\] \[=i\,\left( \frac{1-i}{1-i} \right)+\frac{{{i}^{2}}(1-i)}{(1-i)}\]\[=i+{{i}^{2}}=i-1\].You need to login to perform this action.
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