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  3. CEE Kerala Engineering

CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2005

  • question_answer
    \[{{i}^{2}}+{{i}^{4}}+{{i}^{6}}+...\]upto\[(2k+1)\]terms,\[k\in N\]is:

    A)  0                                            

    B)  1

    C)  \[-1\]                   

    D)         k

    E)  \[k+1\]

    Correct Answer: C

    Solution :

    \[{{i}^{2}}+{{i}^{4}}+{{i}^{6}}+...\]upto\[(2k+1)\]terms \[=\frac{{{i}^{2}}[1-{{({{i}^{2}})}^{2k+1}}]}{1-{{i}^{2}}}=\frac{-1[1-{{(-1)}^{2k+1}}]}{1+1}\]                 \[=\frac{-1[1-(-1)]}{2}=-1\]


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