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JEE Main & Advanced JEE Main Online Paper (Held On 08 April 2018)

  • question_answer
    Let A be the sum of the first 20 terms and B be the sum of the first 40 terms of the series\[{{1}^{2}}+2\cdot {{2}^{2}}+{{3}^{2}}+2\cdot {{4}^{2}}+{{5}^{2}}+2\cdot {{6}^{2}}+......\]. If\[B-2A=100\lambda \], then \[\lambda \] is equal to:                                            [JEE Main Online 08-04-2018]

    A)  464                            

    B)  496

    C)  232                            

    D)  248

    Correct Answer: D

    Solution :

    \[A={{1}^{2}}+{{2}^{2}}+{{3}^{2}}+......+{{20}^{2}}\]                 \[{{2}^{2}}+{{4}^{2}}+....+{{20}^{2}}\] \[\therefore \]  \[A=\Sigma {{20}^{2}}+4\cdot \Sigma {{10}^{2}}\] \[\text{II}{{\text{I}}_{g}}B=\Sigma {{40}^{2}}+4\cdot \Sigma {{20}^{2}}\] \[B-2A=\Sigma {{40}^{2}}+2\cdot \Sigma {{20}^{2}}-8.\Sigma {{10}^{2}}\] \[=\frac{40\times 41\times 81}{6}+2\times \frac{20\times 21\times 41}{6}\] \[-8\times \frac{10\times 11\times 21}{6}\] \[=\frac{40}{6}(41\times 81+21\times 41-22\times 21)\] \[=\frac{40}{6}(4100+82-462)\] \[=\frac{40}{6}(3720)=24800\] \[\therefore \lambda =248\]


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