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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2004

  • question_answer
    If the sum of n terms of the series\[{{2}^{3}}+{{4}^{3}}+\]\[{{6}^{3}}\] + ... is 3528, then n equals to:

    A)  10                                         

    B)  7

    C)  8                            

    D)         9

    E)  6

    Correct Answer: E

    Solution :

    Given series is \[{{2}^{3}}+{{4}^{3}}+{{6}^{3}}+...\] \[\therefore \]  \[{{T}_{n}}={{(2n)}^{3}}=8{{n}^{3}}\] \[\therefore \]  \[\Sigma {{T}_{n}}=8\Sigma {{n}^{3}}=\frac{8{{n}^{2}}{{(n+1)}^{2}}}{4}\] \[\Rightarrow \]               \[3528=2{{n}^{2}}{{(n+1)}^{2}}\]                    (given) \[\Rightarrow \]               \[n=6\]


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