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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2011

  • question_answer
    The value of\[i-{{i}^{2}}+{{i}^{3}}-{{i}^{4}}+....-{{i}^{100}}\]is equal to

    A)  \[i\]                                      

    B)  \[-i\]

    C)  \[1-i\]                  

    D)         \[1+i\]

    E)  0

    Correct Answer: E

    Solution :

    \[i-{{i}^{2}}+{{i}^{3}}-{{i}^{4}}+...-{{i}^{100}}\] This form a GP with common ratio\[(r=-1)\]. \[{{S}_{n}}=\frac{i\{1-{{(-i)}^{100}}\}}{1-(-i)}\]                 \[=\frac{i(1-{{i}^{100}})}{1+i}\]                 \[=\frac{i(1-1)}{1+i}\]                     \[(\because {{i}^{100}}=1)\] \[\therefore \]  \[{{S}_{n}}=0\]


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