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.
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