CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2001

  • question_answer
    The sum of the series \[1+2.2+{{3.2}^{2}}+{{4.2}^{3}}+{{5.2}^{4}}+...+{{100.2}^{99}}\]is:

    A)  \[{{99.2}^{100}}\]           

    B)         \[{{100.2}^{100}}\]

    C)  \[{{99.2}^{100}}+1\]      

    D)         \[{{1000.2}^{100}}\]

    E)  \[{{100.2}^{100}}-1\]

    Correct Answer: C

    Solution :

    Let \[S=1+2.2+{{3.2}^{2}}+{{4.2}^{3}}+....+{{100.2}^{99}}\]...(i) \[\therefore \] \[2S=1.2+{{2.2}^{2}}+{{3.2}^{3}}+....+{{100.2}^{100}}\] ...(ii) On subtracting Eq. (i) from (ii), we get \[-S=1+1.2+{{1.2}^{2}}+{{3.2}^{3}}+.....+{{1.2}^{99}}\]                                                 \[-{{100.2}^{100}}\] \[\Rightarrow \]               \[-S={{2}^{100}}-1-{{100.2}^{100}}\] \[\Rightarrow \]               \[S={{99.2}^{100}}+1\]


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