2. Solved Papers
  3. BCECE Engineering

BCECE Engineering BCECE Engineering Solved Paper-2013

  • question_answer
    If \[1+\lambda +{{\lambda }^{2}}+...+{{\lambda }^{n}}=(1+\lambda )+(1+{{\lambda }^{2}})(1+{{\lambda }^{4}})\]\[(1+{{\lambda }^{8}})(1+{{\lambda }^{16}}),\] then the value of n is (where,\[n\in N\])

    A)  32                         

    B)  16                         

    C)         31                  

    D)         15

    Correct Answer: C

    Solution :

    \[LHS=\frac{1(1-{{\lambda }^{n+1}})}{(1-\lambda )}=\left( \frac{1-{{\lambda }^{n+1}}}{1-\lambda } \right)\] and        \[RHS=(1+\lambda )(1+{{\lambda }^{2}})(1+{{\lambda }^{4}})\]                                 \[(1+{{\lambda }^{8}})(1+{{\lambda }^{16}})\] \[=\frac{[(1-\lambda )(1+\lambda )(1+{{\lambda }^{2}})(1+{{\lambda }^{4}})(1+{{\lambda }^{8}})(1+{{\lambda }^{16}})]}{(1-\lambda )}\]             \[=\frac{(1-{{\lambda }^{32}})}{(1-\lambda )}\] \[\Rightarrow \]               \[\frac{1-{{\lambda }^{n+1}}}{1-\lambda }=\frac{1-{{\lambda }^{32}}}{1-\lambda }\] \[\Rightarrow \]               \[1-{{\lambda }^{n+1}}=1-{{\lambda }^{32}}\] \[\Rightarrow \]               \[n+1=32\] \[\therefore \]  \[n=31\]


You need to login to perform this action.
You will be redirected in 3 sec spinner