CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2000

  • question_answer
    If\[{{(i)}^{2}}=-1,({{i}^{2}})+{{(i)}^{4}}+{{(i)}^{6}}+...\]to\[(2n+1)\]terms is equal to:

    A)  \[-1\]  

    B)                                         1            

    C)         0                            

    D)                         2

    E)         \[-2\]

    Correct Answer: A

    Solution :

    \[{{(i)}^{2}}+{{(i)}^{4}}+{{(i)}^{6}}+....+(2n+1)th\text{ }term\] \[=-1+1-1+1...(-1)=-1\]


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