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BCECE Engineering BCECE Engineering Solved Paper-2006

  • question_answer
    The value of \[x\] for which the equation \[1+r+{{r}^{2}}+...+{{r}^{x}}\] \[=(1+r)(1+{{r}^{2}})(1+{{r}^{4}})(1+{{r}^{8}})\] holds is:

    A)  12                         

    B)         13                         

    C)  14                         

    D)         15

    Correct Answer: D

    Solution :

    \[(1+r)(1+{{r}^{2}})(1+{{r}^{4}})(1+{{r}^{8}})\] \[=(1+r+{{r}^{2}}+{{r}^{3}})(1+{{r}^{4}})(1+{{r}^{8}})\] \[=(1+r+{{r}^{2}}+{{r}^{3}}+{{r}^{4}}+{{r}^{5}}+{{r}^{6}}+{{r}^{7}})(1+{{r}^{8}})\] \[=1+r+{{r}^{2}}+....+{{r}^{11}}+{{r}^{12}}+{{r}^{13}}+{{r}^{14}}+{{r}^{15}}\] Thus for \[x=15,\,\]the equation                 \[1+r+{{r}^{2}}+....+{{r}^{x}}\]                 \[=(1+r)(1+{{r}^{2}})(1+{{r}^{4}})(1+{{r}^{8}})\]holds.


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